Mathcounts Preparatory Packet

Written for the Suzanne Mathcounts 2011-2012 Team

*ALL problems shown here are previous Mathcounts problems, and therefore are most likely very similar to the types of problems you are likely to encounter on all levels of the Mathcounts program.

COUNTING

1. (Sprint) What is the probability that a randomly selected factor of 35 27 is a multiple of 35 14 ? Express your answer as a common fraction.a. 169/784b. 196/729c. ¼d. 25/81e. 225/676

2. (Team) Two numbers, a and b, are randomly selected without replacement from the set {2,3,4,5,6} . What is the probability that the fraction a/b is less than 1and can be expressed as a terminating decimal? Express your answer as a common fraction in lowest terms.a. 1/5b. 3/10c. 2/7d. 3/14e. 2/9

3. (Team) A bag contains blue and green marbles. If 8 green marbles are removed from the bag, the probability of drawing a green marble from the remaining marbles would be 2/5 . If instead 8 blue marbles are added to the bag, the probability of drawing a blue marble would be 17/31 . What was the number of blue marbles in the bag before any changes were made?a. 8b. 7c. 12d. 9e. 6

4. (Sprint) A jar contains 6 red balls, 7 green balls and 9 yellow balls. Danny draws a ball at random from the jar and does not return it. Celine draws another ball at random from the jar. What is the probability that Danny's ball is red and Celine's ball is green? Give your answer as a common fraction in lowest terms.a. 0b. 2/11c. 1/11d. -1/11e. 3/11

5. (Target) We randomly select 6 prime numbers without replacement from the first 17 prime numbers. What is the probability that the sum of the four selected numbers is odd? Express your answer as a common fraction in lowest terms.a. 6/17b. 5/17c. 7/17d. 4/17

e. 8/17

6. (Team) Randomly select a number x such that 0 < x < 58 . What is the probability that x is the sum of two perfect squares? Express your answer as a fraction in lowest terms.a. 25/57b. 26/57c. 8/19d. 29/57e. 23/57

7. (Target) A point is selected inside a rectangle with sides 7 and 11 units. What is the probability that the point is more than 2 units from each vertex. Use Pi=3.14 . Express your answer as a decimal to the nearest hundredth.a. 0.83b. 0.82c. 0.84d. 0.87e. 0.81

8. (Sprint) A three-digit positive number is picked at random. What is the probability that the sum of its digit is equal to 23? Give your answer as a common fraction in lowest terms.a. 16/899b. 1/60c. 17/900d. 1/50e. 9/400

9. (Team) Two distinct points are selected at random from the 10-point set (1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(2,5). What is the probability that the distance between them on a Cartesian grid is a rational number? Express your answer as a decimal to the nearest hundredth.a. 0.57b. 0.58c. 0.56d. 0.53e. 0.59

10. (Target) Find the probability that four randomly selected points of the geoboard shown will be vertices of a square who sides have integer length. Give your answer as a common fraction. The distance between two adjacent vertical or horizontal dots is 1. (The geoboard is a 6x6 grid of dots) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

a. 55/58904b. 55/58903c. 55/58908d. 1/1071e. 55/58902

SOLUTIONS1. c. ¼

Notice that 3527 = 527 * 727. For all multiples of 3514, you need the powers of 5 and 7 to be greater than 14. Therefore, if the numbers are both factors of 3527 and multiples of 3514, then they must be of the form 5a * 7b, where a and b are integers from 14 to 27. Therefore, there are 142 numbers satisfying our requirements, and 282 total factors of 3527, giving us a total of a probability of 142/282 = ¼.

2. b. 3/10

There are 5 choose 2, or 20 such pairs. Note that b has to be either 4, 5, or 6, and that a<b. Therefore, our only answers are {2,4} {2,5} {3,4} {3,5} {4,5} and {3,6}. We have 6/20, or 3/10 probability.

3. d. 9

Looking at the fractions given, we especially notice 17/31. Let us assume that the total number of marbles after 8 blue marbles have been added is 31, and see if it works. Then, here, there are 17 blue marbles and 14 green marbles. Originally, then, there must have been 9 blue marbles and 14 green marbles. Removing 8 green marbles gives us 6 greens, and the new total is 15. Therefore, the probability of drawing a green marble after removing 8 of them is 6/15, or 2/5. It looks like our original assumption works! Therefore, our answer is 14 green marbles and NINE BLUE MARBLES.

4. c. 1/11

There are originally a total of 6+7+9 = 22 balls. Danny’s probability of drawing a red ball is 6/22 = 3/11. Then, there are 21 balls left, and Celine’s probability of drawing a green ball is 7/21 = 1/3. Multiplying these together gives us 1/11.

5. a. 6/17

Remember that all primes are odd except 2. Therefore, if you’re adding six prime numbers together and you get an odd number, then one of them has to be 2 by parity (odd/even). The other five may be picked any way you like. Therefore, we have 16 choose 5 = 4368 possible combinations, and 17 choose 6 = 12376 total choices. Our answer is 4368/12376 = 6/17.

6. b. 26/57

This problem can pretty much be brute forced. Finding all possible sums of squares below 58 gives us the list {1,2,4,5,8,9,10,13,16,17,18,20,25,26,29,32,34,36,37,40,41,45,49,50,52,53}, which is a list of 26 numbers. We have a total of 57 numbers between 0 and 58, so our answer is 26/57.

7. c. 0.84

Draw out the areas in which the point cannot be. You’ll notice that these are four quarter-circles of radius 2 located at the vertexes. The probability is 1 minus the probability of the point being in one of these quarter-circles. The total area is 4pi, and the rectangle’s area is 77. 1 – 4pi/77 = about 0.84.

8. b. 1/60

Simply find all possible 3 digit numbers satisfying this condition (there aren’t that many!). The possible set is {995, 959, 599, 986, 968, 896, 869, 689, 698, 977, 797, 779, 887, 788, 878}. There are 15 numbers in this list, and 100 through 999 or 900 three digit numbers. Therefore, the answer is 15/900 = 1/60.

9. c. 0.56

It’s rather easy to see that if the two points’ distance is a rational number, then they must be on either a horizontal or vertical line. The two can either be in the bottom row (5 choose 2 = 10), top row (5 choose 2 = 10), or along a vertical line (5 lines => 5), giving a total of 25 possibilities. There are 10 choose 2 = 45 pairs, so 25/45 is approximately 0.56.

10. d. 1/1071

We do casework by counting the integer side lengths of the squares: 1, 2, 3, 4, 5, or 6. Counting the number of places for each square individually, we quickly see that there are exactly (6-i)^2 ways to place the square with side length i; in other words, there are 25 places for the unit square, 16 for the side length of 2 square, and so on. One might argue that the 5x5 square can be determined using a 3x4 right triangle to produce a hypotenuse of length 5, which becomes one of the sides of the square; however, upon inspection, you’ll see that you’d require a 7x7 grid to place a square like that. So no, there is no extra way to place the 5x5 square. The total number of squares is 12 + 22 + ... + 52 = 55, and there are a total of 36 choose 4 possible sets of four points = 58905 sets of four points. 55/58905 produces our answer of 1/1071.

*Note: Notice that this problem involved not only combinatorics (counting), but also geometric and algebraic elements. Problems at higher levels of Mathcounts will involve combination of several different types of math, so to be successful you have to be proficient in all areas.

ALGEBRA

1. (Countdown) If 55 * 55 = 5 * 5 * n2 , find n .a. 11b. 22c. 5d. 33e. 55

2. (Sprint) LetS1={1,2,3,4,5,6,7},S2={8,9,10,11,12,13,14},S3={15,16,17,18,19,20,21},S4={22,23,24,25,26,27,28} and so on.What is the sum of the 7 numbers in S69?a. 3360b. 3361c. 3362d. 3357e. 3363

3. (Sprint) The chickens and pigs in Farmer McCoy's barn have a total of 48 heads and 134 legs. How many pigs are in the barn?a. 18b. 17c. 22d. 16e. 19

4. (Sprint) The units digit of a six-digit number is 2 and is removed, leaving a five-digit number. The removed units digit 2 is then placed at the far left of the five-digit number, making a new six-digit number. If the new number is 1/3 of the original number, what is the original number?a. 857142b. 857152c. 857162d. 857112e. 857172

5. (Target) What is the 10-th term in a geometric sequence if the first term is 32 and the 14-th term is 1/256 ?a. 1/32b. 1/8c. 1/16d. ¼e. none of these are correct

6. (Target) What is the sum of all positive integer values of n such (n +25)/n is an integer?a. 31b. 42c. 36d. 60e. 39

7. (Team) The number 135,135 is the product of several consecutive positive odd numbers. What is the greatest of these numbers?a. 12b. 11c. 13d. 16e. 10

8. (Countdown) Solve for x : 1/(10^1) + 1/(10^2) + 1/(10^3) + 1/(10^4) + 1/(10^5) + 1/(10^6) + 1/(10^7) + 1/(10^8)= x/(10^8).a. 1111b. 11111c. 111111d. 11111111e. 1111111

9. (Sprint) What is the sum of the positive whole number divisors of 90?a. 233b. 232c. 237d. 231e. 234

10. (Target) A number can be written in the form (m+11)(n+22) , where m and n are two-digit numbers created by using each of the digits in this set {2, 3, 4, 7} exactly once. What is the smallest such number?a. 2064b. 2063c. 2068d. 2062e. 2065

SOLUTIONS1. a. 11

Simple factorization of 55 x 55 gives us 5 x 5 x 112.

2. a. 3360

Note that every set Si’s last member is equal to 7i. Therefore, the last member in S69 is 69*7 = 483. Here, instead of adding up all the numbers, we employ a sly trick. Note that all the units digits in the answers are different. Therefore, we add up all the units digits of the members in the set instead! 3 + 2 + 1 + 0 + 9 + 8 + 7 = 30, whose units digit is 0. We find the answer choice whose units digit is 0: 3360.

3. e. 19

Using a simple system of equations (2c + 4p = 134, c + p = 48), we may solve this. Doubling the latter equation and subtracting it from the former gives 2p = 38, or p = 19.

4. a. 857142

Let x be the five digit part. Then, 10*x + 2 = (200000 + x)3. Solving for x gives us 10x + 2 = 600000 + 3x, 7x = 599998, and x = 85714. Therefore, our answer is 857142.

5. c. 1/16

Using the simple geometric term formula, a = 32 and ar13 = 1/256 or 2-8. Therefore, r = .5, and ar9 = 32/512 = 1/16.

6. a. 31

(n+25)/n = 1 + (25/n), so n divides 25. All possible values of n include 1, 5, and 25. Therefore, our sum is 31.

7. c. 13

Notice that 135135 = 135*1001 = 335*11*7*13, which fits perfectly into 3, 5, 7, 9, 11, and 13. Therefore, our answer is 13.

8. d. 11111111

Multiplying through by 108 gives us 107+106+105+104+103+102+10+1=x, which is 11111111.(note that this is a countdown problem, and so should be done rather quickly and with some basis on intuition.)

9. e. 234

90 factors into 2*32*5. Using the sum-of-factors formula, we get (1+2)(1+3+9)(1+5) = 3*13*6 = 234.

10. e. 2065

Note that for m and n to be optimally small, 2 and 3 should be the tens digits and 4 and 7 should be the units digits. With these guidelines, we also notice that for the product (m+11)(n+22) to be minimal, the two terms should be as different as possible (as close as possible for maximum). Therefore, we let m be very small, and n be very large; in other words, m = 24 and n = 37. Then, the resulting product is 2065.

Shrewd use of the units digits in m and n show us that the products’ units digit is either 8 or 5 (8: m’s unit digit is 7 and n’s is 4; 5: m’s unit digit is 4 and n’s unit digit is 7). Therefore, choices a, b, and d can be eliminated. Since we already know 2065 is attainable, and 2068 is greater than 2065, our answer is e. 2065.

*Note: The usage of units digits here is very powerful, allowing us to solve the problem without proving that 2065 is the least. However, note that the only reason it was possible to solve the problem so easily was by using the advantages of multiple choice questions to our advantage by eliminating all impossible answers. If this were a short answer question, we would be forced to resort to brute force or usage of the Rearrangement Inequality, which is a tad bit too powerful for usage on a Mathcounts problem. Look it up if you’re interested!