AoPS:

AoPS:Introduction to

Counting & Probability

Chapter 1

Counting is Arithmetic

Counting Lists of Numbers

Problem1.1

How many #s are in the list

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18?

Obviously there are 18 numbers. That was pretty easy. The counting was done for us!

Problem 1.2

How many #s are in the list

7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,

23,24,25,26,27,28,29?

In other words, how many #s are there between

7 and 29 inclusive? (include 7 & 29 in the count)

Solution

A clever way to approach this problem is to convert it to a problem like problem 1.1, by subtracting 6 from every # in the list:


7 8 9 … 29

-6 -6 -6 -6

1 2 3 … 23


7 8 9 … 29 -6 -6 -6 -6 1 2 3 … 23

You may notice that we found that there are 29-7+1 = 23 #s from 7 to 29, inclusive.

Concept:

Given 2 positive #s, a and b, with b > a, find a formula for how many #s there are between a and b inclusive.

Concept:


We can subtract a-1 from our list of #s from a to b to get a list of #s starting at 1:

Concept:



a a+1 a+2 . . . b

-(a-1) -(a-1) -(a-1) . . . -(a-1)

1 2 3 . . . b – a + 1

Concept:



a a+1 a+2 . . . b

-(a-1) -(a-1) -(a-1) . . . -(a-1)

1 2 3 . . . b – a + 1

Our new list of #s has b – a + 1 numbers in it.

Problem 1.3

How many multiples of 3 are between 62 and 215?

Problem 1.3


We see that 62/3 = 20 2/3, so the simplest multiple of 3 is 3 X 21 = 63. Similarly, 215/3 = 71 2/3, so the largest multiple of 3 is 3 X 71 = 213.

Problem 1.3



So our list is 63,66,69, …,213.

Divide it by 3 to convert it to a list we know how to count:

Problem 1.3



So our list is 63,66,69, …,213.

Divide it by 3 to convert it to a list we know how to count:

21, 22, 23, . . . , 71.

We know how to count this list! Subtracting 20 from each number in the list gives

1, 2, 3, . . . , 51

We know how to count this list! Subtracting 20 from each number in the list gives

1, 2, 3, . . . , 51

DO NOT BE TEMPTED TO DO THIS: 215 – 62 = 153 = 51 3 3 IT DOESN’T ALWAYS WORK! See the next problem

Problem 1.4:

How many multiples of 10 are between 9 & 101?

How many multiples of 10 are between 11 & 103?

We know that 101-9 = 103 – 11 = 92, so shouldn’t you get the same answers? Why aren’t they the same?

Problem 1.4:

List 1: the multiples of 10 are 10, 20, 30, …, 100,

so there are 10 multiples.

Problem 1.4:

List 1: the multiples of 10 are 10, 20, 30, …, 100,


List 2: the multiples of 10 are 20, 30, …, 100,


Problem 1.4:

List 1: the multiples of 10 are 10, 20, 30, …, 100,so there are 10 multiples.

List 2: the multiples of 10 are 20, 30, …, 100, so there are 9 multiples. The shortcut doesn’t work: 101 – 9 = 103 – 11 = 92 = 9.2 10 10 10So how would you know whether the answer is 9 or 10?

Problem 1.5

How many 4-digit numbers are perfect cubes?

Solution:


The smallest 4-digit cube is 1000 = 103

Solution:


The smallest 4-digit cube is 1000 = 103

The largest 4-digit perfect cube is a little harder to find and requires a little experimentation. Start by noting 203 = 8000. By trial & error,

213 = 9261

223 = 10,648

So 9261 = 213 is the largest 4-digit cube & the list is 1000, . . . , 9261.

Solution:

There is a much better way that we can write this list:

103, 113, 123, . . . , 203, 213

So the number of #s in the list is the same as

10, 11, 12, . . . , 20, 21

and that means there are 12 numbers in the list!

Now it’s your turn!

1. How many numbers are in the list

36, 37, 38, …, 92, 93?


4, 6, 8, . . . , 128, 130?


-33, -28, -23, …, 52, 57?

Solutions:

1. 58

2. Dividing each member of the list by 2, we get 2, 3, 4, …, 64,65, and then subtracting 1, we get 1,2,3, …,63,64, so there are 64 numbers.

3. We could add 3 to each member in the list to get -30,-25,-20,…,55,60, and divide by 5 to get -6,-5,-4,…,11,12. Then using the integer formula, we get 12 – (-6) + 1 = 19.

Try some more!


147, 144, 141, . . . , 42, 39?


3 2/3, 4 1/3, 5, 5 2/3, …, 26 1/3, 27?

6. How many positive multiples of 7 are less than 150?

Solutions:

4. First, reverse the list, then divide by 3 to get

13, 14, …, 48, 49, so 49 – 13 + 1 = 37.

5. First multiply each number by 3 to get

11, 13, 15, …, 79, 81. Then we can subtract 1 and divide by 2 to get 5, 6, 7, …, 39, 40. So we get 40 – 5 + 1 = 36.

6. 7 x 21 = 147 < 150 < 154 = 7 x 22, so 21 positive multiples of 7 are less than 150.

OK, one last time!

7. How many perfect squares are between

50 and 250?

8. How many odd perfect squares are between

5 and 211?

9. How many sets of four consecutive positive integers are there such that the product of the four integers is less than 100,000?

Solutions:7. Since 72 < 50 < 82 and 152 < 250 < 162, the squares

between 50 & 250 are 82,92,102,…,152. So there are 15 – 8 + 1 = 8.

8. Since 12 < 5 < 32 and 132 < 211 < 152, we have the list 32,52,72,…,132, which has the same # of members as 3,5,7,…,13 which = 6.

9. Note that 174 = 83521 < 100,000 < 104,976 =184. Since 17.54 ≈ 16 x 17 x 18 x 19, we check 16 x 17 x 18 x 19 = 93,024. Also 17 x 18 x 19 x 20 = 116,280, so 16 x 17 x 18 x 19 is the largest product of 4 consec. pos. integers which is less than 100,000. So there are 16 sets.

Counting with Addition and Subtraction

Problem 1.6 At Northshore High School there are 12

players on the basketball team. All of the players are taking at least one foreign language class. The school offers only Spanish & French as its foreign language classes. 8 of the players are taking Spanish and 5 of the players are taking both languages. How many players are taking French?

Solution:

The players taking French fall into 2 categories: those who take Spanish and those who don’t. The # of players taking French and Spanish = 5 (given in the problem).

Solution:The players taking French fall into 2 categories:

those who take Spanish and those who don’t. The # of players taking French and Spanish = 5 (given in the problem).

Next count the # of players taking French but not Spanish. There are 12 players on the team in total, and 8 of them take Spanish, so there are 12 – 8 = 4 not taking Spanish. Since every player must take at least one language, there are 4 taking French.

Solution:So the # of players taking French is the sum of

the # of players in each of the two categories,

5 + 4 = 9.

There is another way to solve this problem:

Draw a Venn Diagram. Use a Venn Diagram whenever you wish to count things or people which occur in two or three overlapping groups.

See the next page.

Place points in the circles to represent the players. A point that is in the French circle that is not in the Spanish circle represents one player taking French but not Spanish.

French Spanish

A point in the region that is in both circles represents a player taking both languages.

French Spanish

A player taking Spanish but not French is represented by a point inside the Spanish circle but not French one.

French Spanish

Finally a point placed outside both circles represents a player who is in neither class.

French Spanish

Now we can use the diagram to solve the problem. Put 5 points in the intersection of both circles because there are 5 players in both classes.

French Spanish

Now, since there are 8 players taking Spanish, and 5 points are already inside the Spanish circle on the right, there must be 3 more points inside the Spanish circle not in the French circle. Add 3.

French Spanish

Since we have 12 total points and we know there aren’t any outside both circles, there must be 4 left inside the French circle but not inside the Spanish circle so add 4 points.

French Spanish

So now we can just read off the answer – there are 9 points inside the French circle on the left.

French Spanish

4 5 3

Problem 1.7

There are 27 cats at the pound. 14 of them are short-haired. 11 of them are kittens. 5 of them are long-haired adult cats. How many of them are short-haired kittens?

Solution:

Draw a Venn Diagram, with one circle for cats with short hair and one circle for cats which arekittens.

Which #s do we want to place in the regions? Since 5 cats don’t have short hair & are not kittens, we know there are 5 cats outside both circles.

Short Hair Kittens

5

At this point we can’t immediately fill any of the other numbers, because none of our #s corresponds exactly to a region of the diagram. For example, we know there are 11 kittens, but there’s no single region of the diagram that corresponds to “kittens”: there’s a region for “short-haired kittens” and a region for “long-haired kittens.” So were going to have to use a little bit of thought. (Be very careful here)

The part of the right circle that does not intersect with “short hair” must represent “long-haired kittens.”

Short Hair Kittens

5

Short Hair Kittens

5

This is theregion that representslong-hairedkittens.

This regionrepresentsshort-hairkittens.

Introduce a variable x. Call the # of cats in one of the regions inside the circles x and try to find other regions in terms of x. Let the # of “short-haired kittens” be x.

Short Hair Kittens

5

Short Hair Kittens

5

x

Since there are a total of 14 short-haired cats, and x of them are kittens, we know that 14 – x of them are not kittens. Then we have 11 – x kittens that are not short-haired.

Short Hair Kittens

5

Short Hair Kittens

5

x14 - x 11 - x

There’s one more piece of information that we haven’t used yet: the total # of cats = 27. So everything must add up to 27: (14 – x) + (11 – x) + x + 5 = 27 so x = 3

Short Hair Kittens

5

Short Hair Kittens

5

311 8

You Try!

1. There are 20 cars in my building’s parking lot. All of the cars are red or white. 12 of them are red, 15 of them are 4 door, and 4 of them are 4 door and white. How many of the cars are 4 door and red?

Let the # of red 4-door cars be x. Since there are 12 red cars and 15 4-door cars, the # of red 2-door cars is 12 – x, while the # of white 4-door cars is 15 – x.

Red 4-Door

x12 - x 15 - x

4

The sum of the # of red 4-door cars, red 2-door cars, white 4-door cars, and white 2-door cars is the total # of cars, 20, because each white 4-door car is contained in exactly one of these categories.

Red 4-Door

x12 - x 15 - x

4

Since the number of white 2-doors is 4, we have

x + (12 – x) + (15 – x) + 4 = 20,

which makes x = 11.

Red 4-Door

x12 - x 15 - x

4

Another one!

2. Going back to the 12-person basketball team, all 12 players are taking at least one of biology or chemistry. If 7 players are taking biology and 2 are 2 players are taking both sciences, how many players are taking chemistry?

Going back to the 12-person basketball team, all 12 players are taking at least one of biology or chemistry. If 7 players are taking biology and 2 are 2 players are taking both sciences, how many players are taking chemistry?

Solution: 7 players are taking biology, so 12 – 7 = 5 players are not taking biology, which means 5 players are taking chemistry alone. Since 2 are taking both, 5 + 2 = 7 players taking chemistry.

Problem 3

There are 30 students in Mrs. Taylor’s kindergarten class. If there are twice as many students with blond hair as with blue eyes, 6 students with blond hair and blue eyes, and 3 students with neither blond hair nor blue eyes, how many students have blue eyes?

Solution:

Let the number of blue-eyed students be x, so the # of blond students is 2x. Since the # of blue-eyed blond students is 6, the # of blue-eyed non-blond students is x – 6, while the # of blond non-blue-eyed students is 2x – 6.

Since the # of non-blue-eyed non-blond students is 3, we can add up these four exclusive categories to sum to 30 students in the class. So

(x – 6) + (2x - 6) + 6 + 3 = 30 and x = 11.

Blue-eyed Blond

x - 6 2x - 6x

3

Problem 4

At the Good-dog Obedience School, dogs can learn to do 3 tricks: sit, stay, and roll over. Of the dogs at the school:

50 dogs can sit 17 dogs can sit & stay

29 dogs can stay 12 dogs can stay & roll over

34 dogs can roll over 18 can sit & roll over

9 dogs can do all three 9 dogs can do none

How many dogs are in the school? How many dogs can do exactly 2 tricks?

Sit

Stay Roll Over

There are 9 dogsthat can do all 3 tricks and there are9 dogs that can donone.

9

9

Sit

Stay Roll Over

Since 18 dogs can sit and roll over (and possibly stay) & 9 dogs can sit, stay, & roll over, there are 18 – 9 = 9 dogs that sit and roll over, but not stay.9

9

9

Sit

Stay Roll Over

Using the same reasoning, there are12 – 9 = 3 dogs that can stay and roll over but not sit, and 17 – 9 = 8 dogs that can sit & stay, but not roll over.9

9

9

3

8

Sit

Stay Roll Over

So now we know how many can do multiple tricks, & exactly what tricks they can do. Since 50 dogs can sit, 9 dogs can sit & roll over only, 8 dogs can sit & stay only, & 9 dogs can do all three tricks, the remaining dogs that can’t do multiple tricks can only sit, and there are 50 – 9 – 8 – 9 = 24.

9

9

9

3

8

24

Sit

Stay Roll Over

Using the same reasoning, we find that 29 – 3 – 8 - 9 = 9 dogs can only stay and34 – 9 – 3 – 9 = 13 dogs can only roll over.9

9

9

3

8

24

9 13

Sit

Stay Roll Over

Since 9 dogs can do no tricks, we can add each category in the Venn Diagram to find that there are a total of 9+9+3+8+24+13+9+9= 84 dogs and 8 + 9 + 3 = 20 dogs that can do exactly 2 tricks.

9

9

9

3

8

24

9 13

Problem 5

Every student in my school is in either French or Spanish class or both. Let x be the number of students in French class and y be the number of students in Spanish class, and z be the number os students in both classes. Find an expression in

x, y, and z for how many students there are in my school.

Solution:

Since x people are in French and z people are in both, x – z are only in French. Similarly,

y – z are only in Spanish. Everyone in the school is in either French only, Spanish only, or both, so the total # of people in the school is (x – z) + (y – z) + z = x + y + z.

Counting Multiple Events

You have three shirts and four pairs of pants. How many outfits consisting of one shirt and one pair of pants can you make?

Counting Multiple Events

You have three shirts and four pairs of pants. How many outfits consisting of one shirt and one pair of pants can you make?

Easy: 3 x 4 = 12 outfits.

You could also make a tree diagram, but you already know how to do that!

Concept:

We use multiplication to count a series of independent events.

By independent, we mean that each decision does not depend on the others.

2nd Example

In how many ways can we form a license plate if there are 7 characters, none of which is the letter O, the first of which is a numeral digit (0-9), the second of which is a letter, and the remaining five of which can be either a digit or a letter (but not the letter O)?

Each character is independent of any other. There are 10 choices for the 1st character (0-9), 25 choices for the 2nd character (A-Z except O), and there are 35 choices for each of the other five characters (any digit 0-9 or any letter A-Z, except O.

Since the choices are independent, we have

10 x 25 x 35 x 35 x 35 x 35 x 35 = 10 x 25 x 355

= 13, 130, 468, 750.

Arranging Things

In how many ways can I arrange four different books on a shelf?

There are 4 choices for the 1st book, with 3 books remaining. So there are 3 choices for the 2nd book, with 2 books remaining. Then there are 2 choices for the 3rd book, with 1 book remaining. So we have only 1 choice for the last book.

4 x 3 x 2 x 1 = 24 choices for all four books.

What happens when choices are not independent?

Your math club has 20 members. In how many ways can it select a president, a vice-president, and a treasurer if no member can hold more than one office?

Once a student is chosen for president, he/she is not available to be chosen for the other offices.

We have 20 choices for president, then 19 choices left for vice-president, and last, 18 choices left for treasurer.

Therefore, there are 20 x 19 x 18 = 6840

ways to fill the three offices.

The last 2 problems are examples of permutations.

A permutation occurs whenever we have to choose several items one at a time from a larger groups of items.

In the 1st problem, we are asked to order four different books. In the 2nd problem, we are asked the # of permutations of 3 people out of 20 people, or how to fill 3 different slots from a group of 20 people.

Factorial!

We can order 4 different objects in 4! different ways.

4! = 4 x 3 x 2 x 1 = 24 ways

Important: 0! means the # of ways to arrange 0 objects in a row. There is only one way to arrange zero objects, do nothing. So

0! = 1

Exercises:

1. For each of 8 colors, I have one shirt & one tie of that color. How many shirt-and-tie outfits can I make if I refuse to wear a shirt & tie of the same color?

2. How many license plates consist of 3 letters, followed by 2 even digits, followed by 2 odd digits?

3. In how many ways can I stack 5 books on a shelf?

1. There are 8 options for the shirt and only 7 choices for the tie, or 8 x 7 = 56.

2. There are 26 choices of letters for each of the 1st two spots & 10 choices of digits for each of the next 3, for a total of 262 x 103 = 676,000.

3. There are 26 choices of letters for the 1st 3 spots & 5 choices for each of the last 4 spots

(5 even or odd digits). 262 x 54 = 10, 985, 000.

Exercises

4. Suppose I have 6 different books, 2 of which are math books. In how many ways can I stack my 6 books on a shelf if I want a math book on both ends of the stack?

5. There are 8 sprinters in the Olympic 100-meter finals. The gold medal goes to 1st place, silver to 2nd, and bronze to 3rd. In how many ways can the medals be awarded?

4. Place the math books 1st. We have 2 choices for the bottom book and 1 choice for the top math book. Then we place the other four books in the middle. There are 4 choices for the 1st, 3 for the 2nd, 2 for the 3rd, and only 1 for the 4th. So the total is 2 x 1 x 4 x 3 x 2 x 1 = 48.

5. There are 8 possible sprinters for gold, then 7 left for silver, and last, 6 left for bronze, for

8 x 7 x 6 = 336 ways to award the medals.

Compute each of the following:

6. 9! ÷ 8!

6. 42! ÷ 40!

6. 8! – 7!

6. 9 x 8! = 9

8!

7. 42 x 41 x 40! = 42 x 41 = 1722

40! 1

8. 8 x 7! – 7! = 7!(8 – 1) = 7! X 7 = 5040 x 7 =

35, 280

Permutations

A club has n members, where n is a positive integer. In how many ways can we choose r different officers of the club (where r is a positive integer, and r < n) such that no member holds more than one office?

A club has n members, where n is a positive integer. In how many ways can we choose r different officers of the club (where r is a positive integer, and r < n) such that no member holds more than one office?

There are n choices for the 1st office, n – 1 for the 2nd, n – 2 for the 3rd, and so on. When we get to the rth office, we’ve already chosen r – 1 members for the previous r – 1 offices, so we have n – (r – 1) = n – r + 1.

There is an easier way to write this:

The number of permutations of n objects taken r at a time is

P(n,r) = n!

(n – r)!

P(30,3) = 30! = 30! = 30 x 29 x 28 x 27!

(30-3)! 27! 27!

= 30 x 29 x 28 = 25, 360

Slidell is running a lottery. In the lottery, 25 balls numbered 1 – 25 are placed in a bin. Four balls are drawn one at a time & their #s are recorded. The winning combination consists of the four selected #s in the order they are selected. How many winning combinations are there if:

(a) each ball is discarded after it is removed?

(b) each ball is replaced in the bin after it is removed & before the next ball is drawn?

(a) 25 choices for the 1st, 24 for the 2nd, 23 for

the 3rd, 22 for the 4th.

25 x 24 x 23 x 22 = 303,600

(b) 25 choices for each of the four balls, or

25 x 25 x 25 x 25 = 254 = 390, 625

The difference between these 2 examples is the difference making selections without replacement and with replacement.

Review:

1. How many #s are in the list

2.5, 5.5, 8.5, 11.5, …, 80.5,83.5?

2. How many 3-digit #s are divisible by 7?

3. There are 20 people in the 7th grade school band. 8 of them are left-handed. 15 of them like jazz music. 2 of them are right-handed and dislike jazz music. How many club members are left-handed and like jazz music?

1. Add 0.5, then divide by 3 to get

1,2,3,4,…27,28 so there are 28 numbers.

2. 7 x 14 = 98 < 100 < 105 = 7 x 15 and

7 x 142 = 994 < 1000 < 1001= 7 x 143. That means the list of 3-digit #s divisible by 7 is 105,112,…,994, and when we divide by 7, we get the list 15,16,17,…,141,142 which has 142 – 15 + 1 = 128 numbers.

3. Let x = left-handed jazz lovers, then 8 – x = left-handed people who dislike jazz and 15 – x jazz lovers are right-handed. Since the # of righty jazz dislikers = 2 and the total # of members of the club = 20, we can add these four exclusive regions to get x + (8 – x) +

(15 - x) + 2 = 20, so x = 5 (lefty jazz lovers).

4. How many 3-letter combinations can be formed if the 2nd letter must be a vowel (a,e,i,o,u) and the 3rd letter must be different from the 1st letter?

5. The local theater has one ticket window. In how many ways can six people line up to buy a ticket? (Source: MATHCOUNTS)

6. Our basketball team has 12 members, each of whom can play any position. In how many ways can we chose a starting lineup consisting of a center, a power forward, a shooting forward, a point guard, and a shooting guard?

4. There are 26 options for the 1st letter, and only 5 options for the 2nd, and only 25 options for the 3rd. This gives 26 x 5 x 25 = 3,250.

5. This is a permutation of 6 people = 6! = 720.

6. This is a permutation of 5 players being chosen in order out of 12, so the answer is

P(12,5) = 12! = 12 x 11 x 10 x 9 x 8 x 7!

(12-5)! 7!

= 12 x 11 x 10 x 9 x 8

= 95, 040

Challenge!

1. How many positive integers less than 500 can be written as the sum of perfect cubes?

Challenge!

1. 73 < 500 < 83, so a3 + b3 must be

1 ≤ a ≤ 7 and 1 ≤ b ≤ 7.

Make a chart of the sum of 2 cubes.

There are 26 such numbers.

13 23 33 43 53 63 73

13 2 9 28 65 126 217 34423 16 35 72 133 224 35133 54 91 152 243 37043 128 189 280 40753 250 341 46863 432 55973 686

Challenge #2

What is the greatest common factor of 5!, 10!, and 15!?

Challenge #2

Since 5! divides 10! and 15! and 5! has no factor larger than 5!, and 5! is a factor of all three, the answer is 5!.

Challenge 3

What is the units digit of sum

1! + 2! + 3! + 4! + 5! + … + 1000!?

Challenge 3

The units digit of 1! is 1, the units digit of 2! is 2, the units digit of 3! is 6, the units digit of 4! = 24 is 4, the units digit of 5! = 120 is 0.

Challenge 3


For all n ≥ 5, n! is a multiple of 5!, which is a multiple of 10, so for all n ≥ 5, the units digit of n! is 0.

Challenge 3


For all n ≥ 5, n! is a multiple of 5!, which is a multiple of 10, so for all n ≥ 5, the units digit of n! is 0.

This means the units digit of the sum is…

1 + 2 + 6 + 4 + 0 + … + 0 = 13, so the answer is 3

Challenge 4

How many of the factorials from 1! to 100! are divisible 9?

Challenge 4

To have a factor of 9, n! must have two factors of 3. The 1st such n for which this is true is 6, since 6! = 6 x 5 x 4 x 3 x 2 x 1.

Challenge 4


Since 9 is a factor of 6! and 6! is a factor n! for all n ≥ 6, the numbers 6!, 7!, 8!, …, 99!, 100! are all divisible by 9.

Challenge 4


Since 9 is a factor of 6! and 6! is a factor n! for all n ≥ 6, the numbers 6!, 7!, 8!, …, 99!, 100! are all divisible by 9.

There are 100 – 6 + 1 = 95 numbers in the list.

Challenge 5

Which integers n satisfy

1 > 1 > 3

2 n 100

and how many such integers are there?

Challenge 5

Multiplying the inequality by 100n, we get

50n > 100 > 3n.

Challenge 5


50n > 100 > 3n.

Since 50n > 100, n > 2 and 100 > 3n, 100/3 > n.

Challenge 5


50n > 100 > 3n.

Since 50n > 100, n > 2 and 100 > 3n, 100/3 > n.

The integers satisfying both inequalities are 3, 4, 5, …, 32, 33, and there 33 – 3 + 1 = 31.

Challenge 6

My classroom has 11 rows of chairs, with 11 chairs in each row. The chairs in each row are numbered from 1 – 11.

(a) How many chairs have odd numbers?

(b) Suppose we replaced 11 with n. Can you find a formula in terms of n for the number of chairs with odd numbers?

Challenge 6

(a) Each row has odd-numbered chairs 1,3,5,7,9,11 for a total of 6 odd-numbered chairs per row. Since there are 11 rows, there are 6 x 11 = 66 chairs with odd numbers.

Challenge 6

(b) There are 2 cases: n is odd, or n is even.

If n is odd, each row has odd-numbered chairs

1,3,5,…, n – 2, n.

Challenge 6



1,3,5,…, n – 2, n. Adding 1 and dividing by 2, we get

1, 2, 3, …, n – 1, n + 1 .

2 2

Challenge 6



1,3,5,…, n – 2, n. Adding 1 and dividing by 2, we get

1, 2, 3, …, n – 1, n + 1 .

2 2

So there are n + 1 odd numbered chairs in each row

2

times n rows, for a total of n(n + 1) .

2

Challenge 6


If n is even, each row has odd-numbered chairs

1,3,5,…, n – 3, n - 1.

Challenge 6



1,3,5,…, n – 3, n - 1. Adding 1 and dividing by 2, we get

1, 2, 3, …, n – 2, n .

2 2

Challenge 6



1,3,5,…, n – 3, n - 1. Adding 1 and dividing by 2, we get

1, 2, 3, …, n – 2, n .

2 2

So there are n odd numbered chairs in each row

2

times n rows, for a total of n(n) = n2 .

2 2

Challenge 7

We connect dots with toothpicks in a grid as shown. If there are 10 horizontal toothpicks in each row and 20 vertical toothpicks in each column, how many total toothpicks are there?

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Challenge 7

Notice that there are 21 rows of dots and 11 columns of dots. Since there are 20 vertical toothpicks in each column, there are 20 x 11

= 220 vertical toothpicks.

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...

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.. .. .

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Challenge 7

Similarly there are 10 horizontal toothpicks in each row and 21 rows, for 21 x 10 = 210 vertical toothpicks.

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Challenge 7

This gives a total of 220 + 210 = 430 toothpicks.

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Fini!

Documents

AoPS: