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Creative Problem Solving Strategies

By Karen Gulya

Table of Contents

Find a Pattern . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 - 3

List the Possibilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 - 5

Try the Possibilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 - 7

Make a Model and Simulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 - 10

Work Backwards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 - 12

Draw a Picture or Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 - 15

Use Logical Reasoning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 - 21

Make a Table or Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 - 23

Break the Problem Down . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .24 – 26

Solve a Simpler Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 - 29

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Find a Pattern

Much of mathematics is based on patterns, so it is important to study patterns in math. We use

patterns to understand our world and to make predictions. We will also use this skill later when we

combine it with the problem-solving skill ‘solve a simpler problem’. Patterns can be as simple or as

complex as you want them to be, and both you and the students can create your own patterns to solve.

List the next three numbers of each pattern or answer the question.

1) 2, 5, 8, 11, ___, ___, ___

2) 0, 2, 5, 9, 14, ___, ___, ___

3) 1, 4, 9, 16, 25, ___, ___, ___

4) 32, 16, 8, 4, ___, ___, ___

5) 1, 2, 0, 3, -1, 4, ___, ___, ___

6) 1, 1, 2, 3, 5, 8, 13, ___, ___, ___

7) Find the pattern and complete the next line of this table.

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1 1

1 2 1

1 3 3 1

1 4 6 4 1

Solutions

1) 2, 5, 8, 11, 14, 17, 20 (add 3)

2) 0, 2, 5, 9, 14, 20, 27, 35 (add 2 then 3 then 4 then 5, etc.)

3) 1, 4, 9, 16, 25, 36, 49, 64 (either add 3 then 5 then 7, etc. or take 12, 22, 32, etc.)

4) 32, 16, 8, 4, 2, 1, ½ or 0.5 (divide by 2 or cut in half)

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5) 1, 2, 0, 3, -1, 4, -2, 5, -3 (one solution is add 1, subtract 2, add 3, subtract 4, etc. Another way to

view this one is a mix of two patterns that alternate 1, 0, -1, -2, -3, with 2, 3, 4, 5)

6) 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 (this is the Fibonacci sequence. After listing the first two terms, we

get the next term by adding the two before it. Did you know this is seen in nature? Look it up!)

7) Find the pattern and complete the next line of this table.

This is Pascal’s Triangle. The outer two terms in each row are always 1. To get any of the inner terms,

add the two numbers diagonally above it (example, 5 = 1+4 and 10 = 4 + 6)

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1 1

1 2 1

1 3 3 1

1 4 6 4 1

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List the Possibilities

Being thorough and complete without wasting time being redundant is also an important skill in

mathematics. Besides promoting organization, learning how to list possibilities efficiently helps us to

organize our thinking for other types of math problems, especially those that require us to choose the

correct fit (such as in factoring polynomials). The key to doing these problems well is to be systematic in

listing the solution.

Problems like these can be created by teachers and also by students.

Note that the students will probably not begin by listing systematically, but they will learn this as they

practice and you model the process.

Sample problem and solution: How many ways can we program a candy machine to recognize 25 cents

if it can take pennies, nickels, dimes, or quarters?

Solution:

Use p for pennies, n for nickels, d for dimes, and q for quarters.

1. 1 q 2. 2 d, 1 n

3. 2 d, 5 p 4. 1 d, 3 n

5. 1 d, 2 n, 5 p 6. 1 d, 1 n, 10 p

7. 1 d, 15 p 8. 5 n

9. 4 n, 5 p 10. 3 n, 10 p

11. 2 n, 15 p 12. 1 n, 20 p

13. 25 p

So, there are thirteen ways to do this.

Note that the way I solved this was to start with the biggest number of quarters, then the biggest

number of dimes, then the biggest number of nickels. I moved them down systematically.

List the possibilities.

1) List all the ways you can make a meal with one meat, one starch, and one vegetable if you have

beef or chicken, rolls or potatoes, and carrots, or green beans.

2) List all the three digit even numbers you can have if the second digit is a 3 and the number is

less than 500.

3) List all the positive integers that divide 72.

4) Five dots are placed on a paper. List all the ways you can connect any two of the dots.

5) By using the letters in the word ‘mathematics’, list all the ways we can make a two letter

combination where one of the letters is a vowel and the other is a consonant. Order matters,

so ‘ma’ is not the same as ‘am’.

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6) List all the ways we can add two positive integers to get 15 where the first number is smaller

than the second one.

Solutions

1) List all the ways you can make a meal with one meat, one starch, and one vegetable if you have beef

or chicken, rolls or potatoes, carrots, or green beans.

Beef-rolls-carrots beef-rolls-green beans beef-potatoes-carrots

Beef-potatoes-green beans chicken-rolls-carrots chicken-rolls-green beans

Chicken-potatoes-carrots chicken-potatoes-green beans

2) List all the three digit even numbers you can have if the second digit is a 3 and the number is less than

500.

The first digit can be 1, 2, 3, or 4

The second digit can be 3

The third digit can be 0, 2, 4, 6, or 8

130, 132, 134, 136, 138, 230, 232, 234, 236, 238, 330, 332, 334, 336, 338, 430, 432, 434, 436, 438

3) List all the positive integers that divide 72.

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

4) Five dots are placed on a paper. List all the ways you can connect any two of the dots

Call the dots A, B, C, D, and E. The combinations would be: ab, ac, ad, ae, bc, bd, be, cd, ce, and de

5) By using the letters in the word ‘mathematics’, list all the ways we can make a two letter combination

where one of the letters is a vowel and the other is a consonant. Order matters, so ‘ma’ is not the same

as ‘am’.

The vowels are a, e, and i. The consonants are m, t, h, c, and s

am, at, ah, ac, as, em, et, eh, ec, es, im, it, ih, ic, is, ma, me, mi, ta, te, ti, ha, he, hi, ca, ce, ci, sa, se, si

6) List all the ways we can add two positive integers to get 15, where the first number is smaller than the

second one.

1 + 14 2 + 13 3 + 12 4 + 11 5 + 10

6 + 9 7 + 8

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Try the Possibilities

Other names for this strategy are ‘trial and error’ or ‘guess and check’. We use this strategy when a

structured or logical approach does not exist. We also use this approach when a structured or logical

approach does exist but we have not yet learned how to utilize the approach. It is normally an

inefficient method but is good to know in case we can’t figure out a problem in a more efficient manner.

Example: if 𝑥2√𝑥 - 32 = 0 and x is a one digit whole number, what number is it?

Solution: if we begin with the number 1, we will find that 1, 2, and 3 won’t work, but the number 4 will.

So, the answer is 4.

Try the possibilities to discover the answer that works.

1) Find a number between 400 and 450 that is divisible by 2, 3, 5, and 7.

2) When the children visited the farm and saw all the goats and chickens, they counted 39 heads

and 108 legs. How many were goats and how many were chickens?

3) Find the smallest prime number that is larger than 300.

4) A perfect number is one in which the number itself is equal to the sum of its factors (excluding

the number itself). One example of a perfect number is 28. The factors of 28 are 1, 2, 4, 7, 14,

and 28. If we remove 28 and add the other factors, we get 1 + 2 + 4 + 7 + 14, which equals 28.

Find the only one digit perfect number.

5) Maria invited three other couples to a dinner party. After she puts the invitations in the

mailbox, she realized she didn’t check to see if the right invitations were placed in the right

envelopes. How many ways could she have put them in the envelopes so that at least one

person got the right invitation?

Solutions

1) Find a number between 400 and 450 that is divisible by 2, 3, 5, and 7.

One solution of many: We need a number between 400 and 450, but this gives us 50 numbers to try.

We can cut this down by looking at the numbers we are testing. In order for 2 to divide the number, it

must be even; and in order for 5 to divide the number, it must end in 0 or 5. Since the number must be

even, though, it must end in 0. Our possibilities now become 400, 410, 420, 430, 440, and 450. Now we

just check to see which of these can be divided by 3 and 7. 3 goes into 420 and 450. Of these two

choices, only 420 is divisible by 7. Therefore, the answer is 420.

2) When the children visited the farm and saw all the goats and chickens, they counted 39 heads and

108 legs. How many were goats and how many were chickens?

We could solve this using a system of equations, but this process may be beyond our students. In that

case, we can try the possibilities until we find the one that fits. There will be many ways to go about

this, but here is one of them. We know we have 39 heads, so the number of goats + the number of

chickens = 39. I will begin by looking at 20 goats and 19 chickens. This will give 118 legs. This is too

many, so we will take the number of goats down to 19. 19 goats and 20 chickens will give 116 legs.

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Next I will try 17 goats and 22 chickens. This gives 112 legs. Now I will try 15 goats and 24 chickens.

This gives 108 legs, which is what we are looking for. The answer is 15 goats and 24 chickens.

3) Find the smallest prime number that is larger than 300.

In order for a number to be prime, the only factors it has are 1 and itself. So, to cut down the

possibilities, we can remove any even number (2 will go into it) and any number that ends in 5 (5 will go

into it). Our first possibility is 301. 3 will not divide this number, but 7 will, so it is not prime. Next, we

will try 303. 3 divides this number, so that is not prime. The next number we will try is 307. 3, 7, 11, 13,

and 17 will not divide this number, so it is prime. Our answer is 307.

4) A perfect number is one in which the number itself is equal to the sum of its factors (excluding the

number itself). One example of a perfect number is 28. The factors of 28 are 1, 2, 4, 7, 14, and 28. If we

remove 28 and add the other factors, we get 1 + 2 + 4 + 7 + 14, which equals 28. Find the only one digit

perfect number.

HERE IS ONE SOLUTION: I WILL BEGIN WITH THE NUMBER 1 UNTIL I find the number that is perfect.

1. The only factor for 1 is 1. If we remove that, the sum is 0. not this one

2. The factors for 2 are 1 and 2. If we remove 2, the sum is 1. not this one either

3. The factors for 3 are 1 and 3. If we remove 3, the sum is 1. nope

4. The factors for 4 are 1, 2, and 4. If we remove 4, the sum is 3. wrong again

5. The factors for 5 are 1 and 5. If we remove 5, the sum is 1. not yet

6. The factors for 6 are 1, 2, 3, and 6. If we remove 6, the sum is 6. Bingo!

So, the answer for this problem is 6.

5) Maria invited three other couples to a dinner party. After she puts the invitations in the mailbox, she

realized she didn’t check to see if the right invitations were placed in the right envelopes. How many

ways could she have put them in the envelopes so that at least one person got the right invitation?

Here is one way to solve the problem. Call the other couples a, b, and c. Now, list all the ways Maria

could have stuffed the envelopes. Here we go!

The way they should have been stuffed: a, b, c

Possibilities

a, b, c (3 right) a, c, b (1 right) b, a, c (1 right) b, c, a (0 right) c, a, b (0 right) c, b, a (1 right)

Of the 6 total possibilities, 4 of them give at least 1 right invitation. So, the answer is 4 ways.

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Make a Model and Conduct a Simulation

We design a model and simulate solutions to problems where an exact answer can only be estimated

and probability is involved or when we do not have the math knowledge needed to solve the problem.

Simulation is based on the premise that the more trials we conduct, the closer we will get to the actual

solution. It involves carefully designing a model that represents the situation, conducting a series of

trials using that model, and averaging the results.

Sample problem: Mrs. Chung’s class has 30 students in it, with 13 boys and 17 girls. If Charles

volunteers to supervise his son and four other randomly selected students on a field trip, how many

other boys can he expect to supervise?

Solution: First, we will design our model. We need to represent the other 29 students in the class (12

boys and 17 girls). To do this, we will use a deck of playing cards where we will take 12 black cards (for

boys) and 17 red cards (for girls). For each trial, we will shuffle the cards and then pick out 4 random

cards. Each time a trial is conducted, all the cards will be placed back in the deck, and it will be shuffled

at least 7 times to ensure a random sample. We will mark ‘b’ for each black card and ‘r’ for each red

card. Then we will total the number of b’s we have. After all the trials have been completed, we will

average the number of ‘B’ cards. This will estimate how many other boys Charles can expect to

supervise. The trials and estimate are displayed here.

After conducting 10 trials, I obtained an average of 1.5. If I round this up, I get 2. This also works with

the trial outcomes, since 5 out of 10 of them produced 2. So, my estimate from the data would be that

Charles can expect to take two other boys with his field trip group, giving a group of his son, 2 other

boys, and 2 girls.

Design a simulation and work through at least five trials to estimate a reasonable answer.

Note: because these are simulations, answers may vary.

1) Your boss calls a lunch meeting but provides lunch boxes for all the employees. These boxes

contain either a turkey sandwich, a roast beef sandwich, or an egg salad sandwich; and either an

apple or a banana. How many boxes should you expect to examine until you find a box with a

roast beef sandwich and a banana?

2) Choco-Puffs has six different prizes in their cereal boxes this year. How many boxes would you

expect to buy before you get all six prizes?

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3) Rodney’s basketball team was down 1 point as the final buzzer sounded. However, he was

fouled at the final buzzer and is allowed to shoot 2 free throws. If he normally makes 2 out of 3

free throws, will his team be able to win the game? What about Tie the game?

4) Joan works as a surgery center valet. The spaces in her lot are numbered and the car keys are

kept in a box with hooks that are numbered to match the space in which the car is parked. One

day she has twelve cars in her lot when she drops the box. All the keys fall off their hooks and

out of the box. If she picks up the keys and randomly puts them on hooks, about how many will

she get in the right place?

Solutions

1) Your boss calls a lunch meeting but provides lunch boxes for all the employees. These boxes contain

either a turkey sandwich, a roast beef sandwich, or an egg salad sandwich; and either an apple or a

banana. How many boxes should you expect to examine until you find a box with a roast beef sandwich

and a banana?

We will begin by designing our model. To simulate the sandwich choices, we will use a die (singular for

dice). If we get a 1 or 2, we will say that is turkey. If we get a 3 or 4, we will say that is roast beef. If we

get a 5 or 6, we will say that is egg salad. Next, we will flip a coin for the fruit choice. If we get heads,

we will say we have an apple. If we get tails, we will say we have a banana. In order to get the box we

want, we will need a 3 or 4 and also a tails (3T or 4T). For each trial, we will keep rolling the die and

flipping the coin until we get one of the two desired results. We will then record how many times we

had to do this to get the desired result.

Remember, this is only sample data. Your data may

be different.

According to my data, it will take 7 boxes before I

find the one I want. In reality, is should take about 6

boxes, which means my average is slightly high,

although it is supported by my data. The more trials

done (for example, if 15 groups of 2 did 10 trials each

for a total of 150 trials), we would get closer to that

correct number of 6.

2) Choco-Puffs has six different prizes in their cereal boxes this year. How many boxes would you expect

to buy before you get all six prizes?

Again, we will begin by designing our model. Since there are six prizes, we can use a die to simulate this

situation. For this one, we want all six prized, so we will need to keep rolling the die until we get all six

of the numbers to appear. We will record each roll of the die until we get all six and then count how

many times we had to roll the die.

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In this sample of five trials, we got an average of 15.4,

which would round to 15 boxes. However, because

this number is near the middle, we may want to

round up to get 16 boxes. This is personal

preference, though, so that is something we could

discuss with our class to gain their perspective on the

situation.

3) Rodney’s basketball team was down 1 point as the final buzzer sounded. However, he was fouled at

the final buzzer and is allowed to shoot 2 free throws. If he normally makes 2 out of 3 free throws, will

his team be able to win the game? What about tie the game? To design this model, we can use a die

once again. Rodney makes 2 out of 3 free throws, so we can let a

roll of 1, 2, 3, or 4 be a success and a roll of 5 or 6 be a failure.

We only need to roll the die twice for each trial and then record

how many points he scored. If he scored 0 points, the team

would lose. If he scored 1 point, the teams would tie. If he

scored 2 points, the team would win.

Out of the ten trials, we found that Rodney’s team will win 6 of

those times. So, the team will probably win. They will tie 3 of

those times, so they may tie. The team will lose only 1 out of 10

times, so it is unlikely that they will lose.

4) Joan works as a surgery center valet. The spaces in her lot are numbered and the car keys are kept in

a box with hooks that are numbered to match the space in which the car is parked. One day she has

twelve cars in her lot when she drops the box. All the keys fall off their hooks and out of the box. If she

picks up the keys and randomly puts them on hooks, about how many will she get in the right place?

To make a model for this problem, we need something that has twelve choices. Since we don’t have a

12-sided die, we can use a deck of cards for this simulation. We will pull out an ace, 2, 3, 4, 5, 6, 7, 8, 9,

10, jack, and queen of one suit to simulate the 12 keys. The ace will stand for key #1, the Jack for key

#11, and the queen for key #12. We will shuffle the deck each time and then list the cards as they come

off the pile. We will compare this solution with what we would see in the key box, where they would be

in order from 1 to 12.

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Work Backwards

We use the strategy of working backwards when we have the answer to a problem or the point at which

we ended. In this case, we can trace our way back to see what was happening at the beginning.

For example, let’s say that Balraj has $143 in his cash register right now. His cash register tape shows he

rang up a $42 order, issued a credit of $28, and rang up another order of $77. How much was in his cash

register when he began?

To solve this problem, we start with the ending number of 143 and then subtract the orders and add the

credit. We need to do the opposite since we are working backwards (we have to undo what we did).

So, we have 143 – 42 + 28 – 77. This gives us the original amount of $52.

Work backwards to find the solution to each problem.

1) If I take my mystery number, multiply it by 2, add 4, divide by 5, and subtract 2, I get 0. What is

my mystery number?

2) Firefighter Stan is hard at work putting out an office fire. He started on a certain rung of a

ladder, went up 3 rungs, down 5 rungs, up 1 rung, down 2 rungs, down 7 rungs, and up 4 rungs.

He is now on the 12th rung. What rung did he start on?

3) Sasha spent one-third of her money on shoes, then another $36 on a haircut and style. After

that, she spent half of what was left on a dress and finally spent $12 on lunch. She how has $30.

How much money did she begin with?

4) Ted and Ned are playing a Dice game that they created. The rule for scoring in this game is as

follows: each person begins with the same number of points. At the end of each round, the

person with the lowest score on his dice wins. At the end of each round, each person deducts

the total number of points on his dice, but the loser also has to deduct the winner’s points as

well. The first five rounds go as follows: Round 1: ted = 5, Ned = 8; Round 2: Ted = 9, Ned = 4;

Round 3: Ted = 7, Ned = 6; Round 4: Ted = 3, Ned = 10; Round 5: Ted = 9, Ned = 5. If ted’s

final score is 168, how many points did each man begin with?

Solutions

1) If I take my mystery number, multiply it by 2, add 4, divide by 5, and subtract 2, I get 0. What is my

mystery number?

Since we are working backwards, we will do the opposite of each step.

0 + 2 = 2 2 x 5 = 10 10 – 4 = 6 6 ÷ 2 = 3. Our mystery number is 3.

2) Firefighter Stan is hard at work putting out an office fire. He started on a certain rung of a ladder,

went up 3 rungs, down 5 rungs, up 1 rung, down 2 rungs, down 7 rungs, and up 4 rungs. He is now on

the 12th rung. What rung did he start on?

Working backwards means beginning on the 12th rung and undoing the steps.

12 – 4 = 8 8 + 7 = 15 15 + 2 = 17 17 – 1 = 16 16 = 5 = 21 21 – 3 = 18 He began on the 18th

rung.

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3) Sasha spent one-third of her money on shoes, then another $36 on a haircut and style. After that, she

spent half of what was left on a dress and finally spent $12 on lunch. She how has $30. How much

money did she begin with?

Let’s begin at the end with $30.

30 + 12 = 42 42 x 2 = 84 84 + 36 = 120 120 is 2/3 of what she started with. 120 = (2/3)x. Now

divide both sides by 2/3. 120 ÷ 2/3 = 120 x 3/2 = 120/1 x 3/2 = 360/2 = 180.

So, Sasha began with $180.

4) Ted and Ned are playing a Dice game that they created. The rule for scoring in this game is as follows:

each person begins with the same number of points. At the end of each round, the person with the

lowest score on his dice wins. At the end of each round, each person deducts the total number of points

on his dice, but the loser also has to deduct the winner’s points as well. The first five rounds go as

follows: Round 1: ted = 5, Ned = 8; Round 2: Ted = 9, Ned = 4; Round 3: Ted = 7, Ned = 6; Round 4:

Ted = 3, Ned = 10; Round 5: Ted = 9, Ned = 5. If ted’s final score is 168, how many points did each man

begin with?

Let begin at the end again! Ted has a final score of 168 points.

He lost round 5, so we need to add both scores (5+9). 168 + 14 = 182.

He won round 4, so we only need to add his score of 3. 182 + 3 = 185.

He lost round 3, so we add both scores (7+6). 185 + 13 = 198.

He lost round 2, so we add both scores (9+4). 198 + 13 = 211.

He won round 1, so we only add his score of 5. 211 + 5 = 216.

Both men started with 216 points.

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Draw a Picture or Diagram

We use a picture or a diagram to solve a problem when there are many pieces of information in a

problem that need to be organized and can best be organized in picture form. For example, look at this

problem: 28 students were surveyed and asked what modes of transportation they used to commute to

school and back. All 28 used either an automobile (A), a bicycle (B) or a cab (C), some of them used two

modes of transportation, and some of them used all three. Use the information below to figure out how

many students used both an automobile and a bicycle.

• A total of 16 people chose automobiles.

• 5 people chose both automobiles and cabs but not

bicycles.

• Of the people who used only one mode of

transportation, 1 used an automobile, 2 used bicycles,

and 4 used cabs.

The greatest number of students used all three modes of

transportation.

Draw a diagram or picture to solve each problem.

1) Carla’s house is 4 miles due north of Sam’s house. Lily’s house is three miles away from Carla’s

house and two miles away from Sam’s house. Show the possible locations for Lily’s house.

2) Draw a triangle. Connect each vertex with the midpoint of the opposite side. How many total

triangles does this produce?

3) Farmer bob wants to build a rectangular pen using 36 feet of fence. One side of the pen will be

the barn. Find the dimensions and area of the largest such pen he can make, provided the

lengths of the sides will be integers.

4) How many different line segments do we need to connect each vertex of a hexagon with every

other vertex?

Solutions

1) Carla’s house is 4 miles due north of Sam’s house. Lily’s house is three miles away from Carla’s house

and two miles away from Sam’s house. Show the possible locations for Lily’s house.

To solve this problem, we will begin by marking Carla’s house and Sam’s house, measuring out a

distance of 4 between them. Then, we will use a compass set to 3 to draw a circle around Carla’s house.

Finally, we will use a compass set to 2 to draw a circle around Sam’s house. These circles will intersect in

two points, which will give the two possible locations of lily’s house.

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2) Draw a triangle. Connect each vertex with the midpoint of the opposite side. How many total

triangles does this produce?

We just follow the directions to draw this series of triangles. Next, we need to count them. There Is 1

large triangle, 6 small triangles, 6 triangles that are exactly half the size of the large triangle, and 3

triangles that use the center vertex and a complete side. This gives a total of 16 triangles.

3) Farmer Bob wants to build a rectangular pen using 36 feet of fence. One side of the pen will be the

barn. Find the dimensions and area of the largest such pen he can make, provided the lengths of the

sides will be integers.

To solve this problem, we will begin by drawing the diagram and then labeling the sides. After that, we

will try the possibilities until we arrive at the correct answer.

The total amount of fencing is 36, so if we let the small sides be x, the large side will be 36 – x – x, or 36-

2x

As we can see from the chart, the dimensions will be 9 feet by 18 feet

with a total area of 162 square feet.

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4) How many different line segments do we need to connect each vertex of a hexagon with every other

vertex?

To solve this problem, we draw line segments while being very careful to count them. We need 5 total

segments from the first vertex, 4 new segments from the second vertex, 3 new segments from the third

vertex, 2 from the fourth, 1 from the fifth, and 0 from the sixth. This gives a total of 15 line segments.

16

Use Logical Reasoning

Logical reasoning . . .

• can also be called matrix logic or deductive reasoning

• is used when we are given clues to a situation that we need to sort through, think about,

rearrange, and put back together again in order to untangle a situation

• moves from generic information to a specific solution

• Sample Problem: Whose Shoes? Johnny Red, Samantha Blue, and Joey Green are wearing red,

blue, and green shoes, but no person is wearing the same color as his or her last name. If the

person wearing the green shoes is Johnny’s cousin, tell what color shoe each person is wearing.

Logical matrix solution

Red Blue Green

Johnny

Red

x * x

Samantha

Blue

x x *

Joey

Green

* x x

X = no

* = yes

Use logical reasoning to solve these problems

1) Brad, Chad, and Tad are taking Mandy, Sandy, and Candy to the Sunday summer concert. From

the clues below, discover which couples will be attending.

• Tad and Mandy are siblings and live on Elm Street.

• Chad is picking up his date in his new red sports car.

• Tad is taller than Candy’s date.

• Sandy and her date will ride their bicycles to the concert.

• Brad’s date lives on Oak Drive.

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2) The Thomas children are 14, 15, and 16 years old. One likes cycling, one likes swimming, and one

likes running. From the clues below, determine each child’s name, age, and favorite sport.

• The oldest likes cycling.

• Shane likes swimming.

• Bridget is older than Katie.

• The 14-yr. old does not like swimming.

3) Adam, Ben, Cami, and Deb are four friends who have different favorite colors. They either like

red, yellow, blue, or green. Use the clues to find each friend’s favorite color.

• Adam likes either yellow or blue.

• Ben likes either yellow or red.

• The girl who likes red is the shortest of the four friends.

• Deb is taller than Adam.

4) If Tom came just before Tim but just after Tina, and if Tessa came just before Tina but just after

Tristan, than what order did they come in?

5) Balraj, Suki, Josue, Ahmed, and Cecilia are all riding the bus to school. From the clues below, tell

what each person’s hobby is and where each person works on campus.

• Suki got on the bus first, followed by the person who likes to race cars and then by the man who

works in the enrollment office.

• Cecilia struck up a conversation with the deli worker who reads as a hobby. In this conversation,

Cecilia admitted to playing video games as her hobby.

• All five people are listed here: Balraj, the man who works at student life, the woman who works

in the gym, Suki, and the man who works in the bookstore.

• The person who works in student life enjoys weight lifting.

• The rock climber, Josue, and the man who works in the bookstore sat on the left side of the bus.

Solutions

1) Brad, Chad, and Tad are taking Mandy, Sandy, and Candy to the Sunday summer concert. From

the clues below, discover which couples will be attending.

Tad and Mandy are siblings and live on Elm Street.

Chad is picking up his date in his new red sports car.

Tad is taller than Candy’s date.

Sandy and her date will ride their bicycles to the concert.

Brad’s date lives on Oak Drive.

From the first clue, we know that Tad and Mandy are not going together. From the third clue, we

know that Tad is not going with Candy. So, Tad is going with Sandy. From the first clue, we know

18

that Mandy lives on Elm Street. From the fifth clue, Brad’s date lives on Oak Drive. This means that

Brad is going with Candy. This also means Chad is going with Mandy.

2) The Thomas children are 14, 15, and 16 years old. One likes cycling, one likes swimming, and one

likes running. From the clues below, determine each child’s name, age, and favorite sport.

The oldest likes cycling.

Shane likes swimming.

Bridget is older than Katie.

The 14-yr. old does not like swimming.

This problem is still simple enough that we might be able to complete it without a matrix.

From clues 1 and 2, we know that Shane is not the oldest, so he is either 14 or 15.

From clue 2 and 4, we find that Shane is not the 14-yr. old. So, Shane is 15 and likes swimming.

From clue 3, we find that Bridget is 16 and Katie is 14.

From clue 1, we find that Bridget likes cycling. So, Bridget is 16 and likes cycling and Katie is 14 and

likes running. (Note: we had to find Katie’s sport in the description of the problem, not in the list of

clues.)

3) Adam, Ben, Cami, and Deb are four friends who have different favorite colors. They either like

red, yellow, blue, or green. Use the clues to find each friend’s favorite color.

Adam likes either yellow or blue.

Ben likes either yellow or red.

The girl who likes red is the shortest of the four friends.

Deb is taller than Adam.

This one does not need a matrix either. From clue 3, it is a girl who likes red (either Cami or Deb).

From clue 2, Ben likes yellow. From clue 1, Adam likes blue. From clues 3 and 4, Deb likes green

which also means that Cami likes red.

4) If Tom came just before Tim but just after Tina, and if Tessa came just before Tina but just after

Tristan, than what order did they come in?

This one will not require a matrix. From the first part of the sentence (up until the comma), we have

that Tina came first, Tom came in the middle, and Tim came last. From the first part of the

sentence, we add that Tessa came before Tina but Tristan came even before her. So, our final order

is Tristan, Tessa, Tina, Tom, and Tim.

5) Balraj, Suki, Josue, Ahmed, and Cecilia are all riding the bus to school. From the clues below, tell

what each person’s hobby is and where each person works on campus.

Suki got on the bus first, followed by the person who likes to race cars and then by the man who

works in the Enrollment Office.

Cecilia struck up a conversation with the Deli worker who reads as a hobby. In this conversation,

Cecilia admitted to playing video games as her hobby.

All five people are listed here: Balraj, the man who works at Student Life, the woman who works in

the Gym, Suki, and the man who works in the Bookstore.

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The person who works in Student Life enjoys weight lifting.

The rock climber, Josue, and the man who works in the Bookstore sat on the left side of the bus.

This one may be complicated enough to require a matrix. From the clues below, here is how I would

make mine:

Clue 1: Suki got on the bus first, followed by the person who likes to race cars and then by the man

who works in the Enrollment Office.

This tells us several things: (1) Suki does not like to race cars and she does not work in Enrollment.

(2) The person who likes to race cars does not work in Enrollment, and (3) it is a man who works in

Enrollment. Here is what our chart would look like after this clue: (Note: x is for no and * is for yes)

Clue 2: Cecilia struck up a conversation with the Deli worker who reads as a hobby. In this

conversation, Cecilia admitted to playing video games as her hobby.

This clue tells us three things: (1) the reader works in the Deli, and (2) Cecilia likes video games, and (3)

Cecilia does not work in the deli. (Note that when we can fill in a ‘yes’ box, we go up and down and

sideways within that part of the matrix and ‘x’ out the other possibilities.)

20

Clue 3: all five people are listed here: Balraj, the man who works at Student Life, the woman who works

in the Gym, Suki, and the man who works in the Bookstore.

This tells us the following: (1) Balraj does not work in Student Life, the Gym, or the Bookstore, (2) a man

works at Student Life, (3) a man works in the Bookstore. After we fill in all of this, we see that Cecilia

must work in the Gym.

Clue 4: The person who works in Student Life enjoys weight lifting.

After we fill this in, we can also deduce that Balraj and Suki are not the weight lifters. We can see that

from the chart. We can also go back to clue 2 to see that Suki likes reading. Next, from the chart, we

can see that Balraj works in Enrollment.

21

Clue 5: The rock climber, Josue, and the man who works in the bookstore sat on the left side of the bus.

From this clue, we find that Josue is not the rock climber and does not work in the bookstore. So, he

works in Student Life and Ahmed works in the Bookstore. We also know that Ahmed is not the rock

climber. Now we look at the chart to see that the man who works in Student Life is also the weight

lifter, hence Josue. This leaves only racing cars for Ahmed which in turn leaves only rock climbing for

Balraj. It is not necessary to complete the bottom part of the chart because now we and answer the

question: Balraj is the rock climber and works in Enrollment. Suki is the reader and works in the Deli.

Josue is the weight lifter and works in Student Life. Ahmed races cars and works at the Bookstore.

Cecilia plays video games and works at the Gym.

22

Make a Table or Chart

We make a table or chart when we have information that needs to be organized before we can move on

and find the solution to a problem. We also use this strategy when we need to find a pattern and

generalize it before we can answer a question. One thing to note here is that the more difficult

strategies will sometimes blend together and also use the less difficult strategies in their solution. Here

is an example of a problem where using a table or chart would be helpful: Find the greatest common

factor of 24, 56, and 108. To do this problem efficiently, we will break each number down into its prime

factors, in order from smallest to largest.

24 2, 2, 2, 3

56 2, 2, 2, 7

108 2, 2, 3, 3, 3

After we list these out, we find all the numbers that appear on all three lists. In this case, they all have a

first 2 and a second 2. They do not all share any other numbers. So, the greatest common factor is 2 x

2, which is 4.

Use a table or chart to solve the following problems

1) What is the ones’ digit of72015 ?

2) How many diagonals does a regular decagon have?

3) Find the smallest positive integer that is divisible by 2, 3, 5, 7, 12, 15, 21, and 24.

4) If 1! = 1, 2! = 1x2, 3! = 1x2x3, 4! = 1x2x3x4, etc., find the ones’ digit of 89!

5) It costs 75 cents to use a country toll road. The toll machine takes exact change only and

accepts nickels, dimes, and quarters. How many combinations of coins must the machine be

programmed to accept?

Solutions

1) What is the ones’ digit of 72015 ?

After we do this enough times to see a pattern emerge, we find that the

one’s digit cycles every 4 times, giving 7, 9, 3, or 1, respectively. So, if we

want the ones’ digit of the 2015th power, we just need to divide 2015 by

4, note the remainder, and apply it to find the solution. 2015 / 4 = 503 r

3. Since the remainder is 3, this fits with the third number in the cycle,

which is 3. Note: If we have no remainder, that is the same as getting a

remainder of 4 and our choice would be 1.

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2) How many diagonals does a regular decagon have?

To solve this problem, we would also need to draw the

figures, connect the diagonals, and count them. That is not

shown here due to space constraints. However, from the

chart to the right, we can see a pattern emerging. The pattern

is 0, 2, 5, 9. This pattern is: +2. +3, +4, +5, etc. To get to 10

sides, we need to continue this pattern out for 4 more times.

Here is our completed pattern: 0, 2, 5, 9, 14, 20, 27, 35. So, a decagon would have 35 diagonals.

3) Find the smallest positive integer that is divisible by 2, 3, 5, 7, 12, 15, 21, and 24.

Before we start making our table or chart, we can look more

closely at the problem to make it simpler for us. (I call this

working smart as opposed to working hard.) Since 2 goes into

12, we can remove 2 as a choice. This is also true for 3 (into

12), 5 (into 15), 7 (into 21) and 12 (into 24). Hmmm . . . Is it ok

to remove 12 since we said 2 and 3 went into them? The

answer is yes, because 2 and 3 also go into 24. Here is our

table and solution.

4) If 1! = 1, 2! = 1x2, 3! = 1x2x3, 4! = 1x2x3x4, etc., find the ones’ digit of 89!

I especially enjoy this problem because there is an interesting

twist in it. If we look at the chart on the right, we see that

when we solve for 5! And beyond, the ones’ digit is always 0.

Therefore, the answer is 0.

5) It costs 75 cents to use a country toll road. The toll machine takes exact change only and accepts

nickels, dimes, and quarters. How many combinations of coins must the machine be programmed to

accept?

This problem uses the ‘list the possibilities’ strategy except

that it is more complex and will require a systematic way to

arrive at the solution. I solved this by (a) using the most

quarters I could and then (b) using the most dimes I could.

The nickels adjusted themselves accordingly. From the table,

we can see that the machine must be programmed to accept

18 combinations.

24

Break the Problem Down into Sub-parts

We break problems down into sub-parts when we are given a lot of information that needs to be

organized and that needs to be solved in steps. This type of problem is complex according to Webb’s

Depth of Knowledge model, so it helps us to develop good organizational and neatness skills, and it

encourages us to be careful in our calculations.

Here is an example: Derek bought 2 notebooks for $1.79 each, 3 pens for $0.89 each, and a pack of

highlighters for $3.49. The tax rate is 8.5%. Derek pays with a $20 bill. The cashier gives in the

following change: a $5 bill, 4 $1 bills, 2 quarters, 1 dime, 1 nickel, and three pennies. Is this the correct

amount of change?

Note: when the problems get complex like this, I would recommend using a calculator. We don’t

want to get bogged down in the computations when we are really learning something else.

1) The first thing we need to find is the amount of Derek’s purchase without the tax: 2 x 1.79 + 3 x

0.89 + 3.49 = 3.58 + 2.67 + 3.49 = 9.74

2) Now we will add in the tax. 9.74 x 0.085 = 0.8279 = 0.83 (rounded) and 9.74 + 0.83 = 10.57.

3) Next, we will find out how much change he should get back: 20.00 – 10.57 = 9.43.

4) Now we will calculate the amount of change Derek did receive: 5 + 4 x 1 + 2 x 0.25 + 0.10 + 0.05

+ 0.03 = 5 + 4 + 0.5 + 0.1 + 0.05 + 0.03 = 9.68.

5) Then we can compare to answer the question: 9.43 is not equal to 9.68, so this is not the

correct amount of change.

6) As an extension activity, you can explore what went wrong. As you investigate further, you will

notice that the cashier gave him $0.25 too much, so he gave Derek an extra quarter.

Now it’s your turn to solve this type of problem.

1) Find all two digit prime numbers whose individual digits are also prime numbers.

2) How many factors of 400 are perfect squares?

3) Joe has 20 feet of lumber. He needs to cut this into three 6 5/8 inch pieces. Every time he

makes a cut, he will lose an additional 1/16 inch of wood. Will he have enough wood? How

much will he have left over or how much more does he need?

4) Find the sum of the whole numbers between 1 and 100 inclusive that are multiples of 5 or 7.

5) Rent-a-ride charges $50 per day plus $0.19 per mile to rent their economy car. You-rent-it

charges $45 per day plus $0.25 per mile for the same car. If Dawn need to rent a car for three

days and travel 325 miles, which rental company will cost her the least?

Solutions

1) Find all two digit prime numbers whose individual digits are also prime numbers.

1) First, we will identify the one digit prime numbers. They are 2, 3, 5, and 7.

25

2) Next, we will list all possible two digit numbers made from these digits: 22, 23, 25, 27, 32, 33,

35, 37, 52, 53, 55, 57, 72, 73, 75, and 77.

3) We can remove all even numbers because they will not be prime (2 goes into them). Our new

list is 23, 25, 27, 33, 35, 37, 53, 55, 57, 73, 75, and 77.

4) Next, we can remove all numbers ending in 5 (5 will go into them) and also all numbers with

double digits (11 will go into them). Our list is now: 23, 27, 37, 53, 57, and 73.

5) Now we can remove all numbers that 3 will go into. We can use a divisibility test or just use our

calculator. Our new list will be: 23, 37, 53, and 73.

6) We will now check to see if these four numbers are prime. We can do that by checking each

number to see if 7 goes into it (we already checked for 2, 3, and 5, so 7 is the only number left.

We do not need to check for 4, 6, 8, 9, or 10 because the prime numbers go into them). 7 does

not go into any of these numbers, so they are all prime.

7) Our final answer is 23, 37, 53, and 73.

2) How many factors of 400 are perfect squares?

1) First, we will list all of the factors of 400. It will be easiest to do this in pairs and stop when we

get to the middle. We have 1,400 2,200 4,100 5,80 8,50 10,40 16,25 20,20.

2) Now we will check to see which ones are perfect squares (the square root button on our

calculators will work fine here). These numbers are 1, 400, 4, 100, 16, and 25. (Did you notice

that when one of the pair was a perfect square then the other one was too? I wonder why that

is . . . )

3) Now we will count these numbers to answer the question. There are 6 factors of 400 that are

perfect squares.

3) Joe has 20 feet of lumber. He needs to cut this into three 6 5/8 inch pieces. Every time he makes a

cut, he will lose an additional 1/16 inch of wood. Will he have enough wood? How much will he have

left over or how much more does he need?

1) First, we will calculate how much wood Joe needs for the pieces. 3 x 6 5/8 = 19 7/8 inches

(again, a calculator that does fractions will work well here.)

2) Next, we will calculate how much will be lost in the cuts. We need to be careful here, because

although there are three pieces, there are only two cuts. 2 x 1/16 = 1/8 inch.

3) Now we will add the amounts together to see how much total wood we need. 19 7/8 + 1/8 =

20.

4) Since Joe has 20 feet of lumber and he needs 2o feet of lumber, then he will have enough but

will have none left over.

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4) Find the sum of the whole numbers between 1 and 100 inclusive that are multiples of 5 or 7.

1) First, we will list the multiples of 5 that are between 1 and 100 inclusive. They are 5, 10, 15, 20,

25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, and 100. Also, their sum is 1050.

2) Next, we will list the multiples of 7 that are between 1 and 100 inclusive. They are 7, 14, 21, 28,

35, 42, 49, 56, 63, 70, 77, 84, 91, and 98. Also, their sum is 735.

3) Now we need to think, because the numbers 35 and 70 appear on both lists. If we are not

careful we will count these numbers twice. So, we will need to add the sums but then subtract

35 and 70 to undo the duplicates.

4) Our answer, then, is 1050 + 735 – 35 – 70, which is 1680.

5) Rent-a-ride charges $50 per day plus $0.19 per mile to rent their economy car. You-rent-it charges

$45 per day plus $0.25 per mile for the same car. If Dawn need to rent a car for three days and travel

325 miles, which rental company will cost her the least?

1) First, we will calculate how much dawn will pay if she uses rent-a-ride. 50 x 3 + 0.19 x 325 =

$211.75.

2) Next, we will calculate how much dawn will pay if she uses you-rent-it. 45 x 3 + 0.25 x 325 =

$216.25.

3) Finally, we compare the two choices and choose the one requiring the smaller amount. $211.75

is less than $216.25, so her lowest cost option is rent-a-ride.

27

Solve a Simpler Problem

When we read a math problem and recognize that it will be very hard to solve it directly, then it is time

to solve a simpler problem. Most of the time, when we solve a simpler problem or a series of simpler

problem, a pattern will emerge that we can extend to solve the larger problem. Another thing that can

happen when we solve a simpler problem is that we will see the path we need to take to get the

solution to the larger problem.

Here is an example: A polygon has 6 times the number of diagonals as it has sides. How many sides

does this polygon have?

We will begin by reviewing the number of diagonals some of the smaller polygons have, finding the

pattern, and extending it out to answer the question. If you remember, we were able to draw out the 3,

4, 5, and 6 sided polygons and count the diagonals they had. From there, we found the pattern and

extended it out. Here is our chart. We can see from the chart that the polygon with six times the

number of diagonals as sides has 15 sides.

Solve a simpler problem to find the solution.

1) Find the sum of the first 50 odd numbers.

2) Find the sum of the whole numbers from 1 to 120.

3) There are 250 pennies laid out in a row and all of them are heads up. Suppose the first person

comes along and flips all of the pennies over. Then the second person flips over every second

penny. The third person flips every third penny. The fourth person flips every fourth penny, the

fifth person flips every fifth penny, and so on, until the 25oth person flips the last penny. At the

end of this, which pennies will be tails up?

4) A pyramid of wooden blocks is stacked up against the wall so that there are 2 blocks in the top

row, 4 in the second row, 6 in the third row, and so on. How many rows are there if the pyramid

contains 756 blocks?

5) How many squares of any size are on an 8x8 chessboard?

Solutions

1) Find the sum of the first 50 odd numbers.

Well, let’s begin small and see if we can find a pattern:

• 1 = 1

• 1 + 3 = 4

• 1 + 3 + 5 = 9

28

• 1 + 3 + 5 + 7 = 16

• 1 + 3 + 5 + 7 + 9 = 25

Wait a minute – we’ve seen the numbers 1, 4, 9, 16, and 25 before – they are the perfect squares. So,

the sum of the first 1 odd number = 1 squared; the sum of the first 2 odd numbers is 2 squared; the sum

of the first three odd numbers is 3 squared; and so on. This means the sum of the first 50 odd numbers

is 5o squared or 50 x 50. This number is 2500.

2) Find the sum of the whole numbers from 1 to 120.

If we try to use a pattern like in the last problem, we won’t get very far. (In the interest of time and

space, I will leave that to you if you want to try it.)

So, let’s look a little deeper and think outside the box. 120 is an even number, so we can pair up

everything. If we pair up (1+2), (3+4), (5+6), that won’t get us anywhere. How about it we pair up (1 +

120), (2 + 119), (3+ 118), and so on? If we do that, each of the pairs has a sum of 121. Now we’re

getting somewhere! How many pair of numbers will we have? 120 ÷ 2 = 60, so we have 60 pairs with a

sum of 121. Our answer will be 60 x 121, which is 7260.

3) There are 250 pennies laid out in a row and all of them are heads up. Suppose the first person comes

along and flips all of the pennies over. Then the second person flips over every second penny. The third

person flips every third penny. The fourth person flips every fourth penny, the fifth person flips every

fifth penny, and so on, until the 25oth person flips the last penny. At the end of this, which pennies will

be tails up? Hmmm . . . 250 is too many to try and simulate a solution to. Let’s try it with 20 pennies

and see what happens.

Begin H H H H H H H H H H H H H H H H H H H H

P1 T T T T T T T T T T T T T T T T T T T T

P2 T H T H T H T H T H T H T H T H T H T H

P3 T H H H T T T H H H T T T H H H T T T H

P4 T H H T T T T T H H T H T H H T T T T T

P5 T H H T H T T T H T T H T H T T T T T H

P6 T H H T H H T T H T T T T H T T T H T H

P7 T H H T H H H T H T T T T T T T T H T H

P8 T H H T H H H H H T T T T T T H T H T H

P9 T H H T H H H H T T T T T T T H T T T H

P10 T H H T H H H H T H T T T T T H T T T T

P11 T H H T H H H H T H H H T T T H T T T T

P12 T H H T H H H H T H H H T T T H T T T T

P13 T H H T H H H H T H H H H T T H T T T T

P14 T H H T H H H H T H H H H H T H T T T T

P15 T H H T H H H H T H H H H H H H T T T T

P16 T H H T H H H H T H H H H H H T T T T T

P17 T H H T H H H H T H H H H H H T H T T T

P18 T H H T H H H H T H H H H H H T H H T T

P19 T H H T H H H H T H H H H H H T H H H T

P20 T H H T H H H H T H H H H H H T H H H H

29

At the end of our simulation, we see that pennies 1, 4, 9, and 16 have tails up. Once again, we have

seen these numbers – they are the squares. So, all the square numbers up to 250 will be tails up. They

are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, and 225.

4) A pyramid of wooden blocks is stacked up against the wall so that there are 2 blocks in the top row, 4

in the second row, 6 in the third row, and so on. How many rows are there if the pyramid contains 756

blocks?

Once again, we will work this row by row and look for a pattern.

• Row 1 = 2

• Row 2 = 2 + 4 = 6

• Row 3 = 2 + 4 + 6 = 12

• Row 4 = 2 + 4 + 6 + 8 = 20

• Row 5 = 2 + 4 + 6 + 8 + 10 = 30 ‘

Now, if we keep going, we will have to do this until we get a total of 756, which could take a long time.

So, we will look back at the row numbers compared to the sum and see if we can make a connection. If

we think about it, every row’s sum equals the row number times (the row number + 1). So, row 1 = 1x2

= 2, row 2 = 2x3 = 6, row 3 = 3x4 = 12, row 4 = 4x5 = 20, row 5 = 5x6 = 30. We can use this to find our

answer. For row x, we would have x(x+1) = 756. This means two numbers right next to each other will

multiply to give us 756 and we want the smallest one. So, let’s look at 756. The Square root of 756 (its

middle for multiplication) is 27.49545417 so we will check for 27 and go down until we get the pair. This

pair is 27 and 28. That means the number we want is the smaller one, which is 27. The pyramid

contains 27 rows.

5) How many squares of any size are on an 8x8 chessboard?

Again, if we tried to do this outright, we might lose our sanity! Let’s look at a 1x1 board, then a 2x2

board, etc. and see if any patterns emerge.

1) A 1 x 1 board has only 1 large square. (1 = 12)

2) A 2 x 2 board had 1 large square and 4 small squares. (4 = 22)

3) A 3 x 3 board had 1 large square, 4 medium squares, and 9 small squares. (9 = 32)

We can already see a pattern emerging here, and it involves the squares again (funny how they keep

popping up!) Here is a summary and then on to our solution:

1x1 = 12, 2x2 = 12 + 22, 3x3 = 12 + 22 + 32, therefore, 8x8 = 12 + 22 + 32 +42 + 52 + 62 +72 + 82 = 1 + 4 + 9 + 16

+ 25 + 36 + 49 + 64 = 204 total squares.