Dec 2015

From now until Feb 2016 we will practice Mathcounts problems. Mathcounts is a team competition.  To participate, your school must register a team.

I’ve chosen problems from the 2014-2015 exercises. Unlike AMC8, Mathcounts are short-answer questions, not multiple choices.  Some even allow using a calculator, and I’ll indicate those with “(calculator)” at the end of the problems.  Mathcounts and AMC8 test similar math skills, though I feel Mathcounts is somewhat harder.

Like before, I’ll group problems based on topics. I’ll post new problems and answers to old problems each Saturday, and I’ll give out prizes on a regular basis. Enjoy!

For instructions on submitting weekly answers, click on the “Logistics” tab.
For previously posted problems, click on the “Archive” tab.
For previous prizes, click on the “News” tab.

Week of Dec 19 – 25. This week we will focus on counting. Click GoogleForm to submit answers by Saturday, Jan 2.

Answers: 33, 20, 56, 55, 60, 78, 20

Dec 19. How many distinct three-letter strings can be formed using three of the five letters in the word SILLY?

Dec 20. In how many ways can the vertices of an equilateral triangle be colored with four available colors? Two colorings are considered the same if they can be obtained from each other by any combination of rotations and reflections.

Dec 21. The Statesville Middle School basketball team has 8 players. If a player can play any position, in how many different ways 5 starting players be selected?

Dec 22. Donald has nine one-day passes to Dizzyworld. He can go alone and use one, or he can take a friend and use two. If he visits every day until he uses all the passes, in how many different ways can he use them? Using two a day for four days and then going alone at the end (2, 2, 2, 2, 1) is different from reversing the order (1, 2, 2, 2, 2).

Dec 23. A soccer ball is a polyheron comprised of 12 pentagons and 20 hexagons. How many vertices does a soccer ball have?

Dec 24There are 13 stations along the Cheshire Railrodad, which runs in a straight line from east to west. A “trip” is defined by its starting and ending stations (regardless of intermediate stops) and must always go westward. How many different trips are possible along the Cheshire Railroad?

Dec 25How many squares can be drawn using only dots in this grid of 16 evenly spaced dots as vertices?

 [Solution from Peter] Here’s a link to Peter Han’s Solution.

Week Dec 12 – 18. This week we will focus on probability. Click GoogleForm to submit answers by Friday night.

Answers: 1/8, 1/100, 5/9, 7/36, 121/282, 4/9, 1/12

These probability problems turned out quite difficult. On multiple problems, I received varying answers. For example, there were five different responses for Dec 18’s question. To improve everybody’s probability skills, I’d like to ask you to email me your solutions, with proper explanations.  I’ll post correct solutions, giving credit to the contributors.   If you got a question wrong and would like to know why, you can email me your question and I will post a response on the webpage, keeping you anonymous.

Each explanation, right or wrong, will count as an extra point on the score sheet. Each question will also count as an extra point on the score sheet. Even if you didn’t submit answers, you are still welcome to contribute to the solutions. 

Dec 12. When rolling two fair, eight-sided dice, each with faces numbered 1 through 8, what is the probability that the two numbers rolled have a sum of 9? Express your answer as a common fraction.  (A common fraction refers to a simplified fraction.)
[Solution from Gyan]  Since each dice has 8 numbers, 1,2,3,4,5,6,7 and 8, the only numbers that can be rolled that have a sum of nine are (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2) and (8,1). In total there are 64 combinations and 8 of them with sum equal nine. So the probability is 8/64 which simplifies to 1/8.

Dec 13. Let P(n) denote the probability that a randomly selected n-digit number contains the digits 42, adjacent and in that order, among its digits. Two 4-digit examples are 3422 and 4205. What is the absolute difference between P(2) and P(3)? Express your answer as a common fraction.
[Solution from Stephen]  There are 90 2-digit numbers, and only one of them 42 contains the digits 42. So P(2)=1/90.  There are 900 3-digit numbers.  If 42 appears on the hundreds and tens positions, there are 10 possibilities on the units position.  If 42 appears on the tens and units positions, there are 9 possibilities on the hundreds position.  So P(3)=19/900.  P(3)-P(2)=19/900-1/90=9/900=1/100.

Dec 14. When rolling two fair, standard dice, what is the probability that the sum of the numbers rolled is a multiple of 3 or 4? Express your answer as a common fraction.
[Solution from Stephen]  The largest sum of 2 dice rolled is 12. We list multiples of 3 and 4 up to 12 and list all the ways that can add to these sums.  For sum equal to 3, we have 1+2, 2+1. For sum 4, we have 1+3, 2+2, 3+1. For sum 6, we have 1+5, 2+4, 3+3, 4+2, 5+1. For sum 8, we have 2+6, 3+5, 4+4, 5+3, 6+2. For sum 9, we have 3+6, 4+5, 5+4, 6+3. For sum 12, we have 6+6. So there are a total of 20 ways to add up to multiples of 3 and 4.  Since there are 6*6=36 ways in total, the probability is 20/36=5/9.

Dec 15. Two fair, six-sided dice are rolled. They are marked so one die has the numbers 1, 3, 5, 7, 9, 11 and the other has the numbers 2, 4, 6, 8, 10, 12. What is the probability that the sum of the numbers rolled is divisible by 5? Express your answer as a common fraction.
[Solution from Gyan] Since 11+12=23, the only sums divisible by 5 are 5, 10, 15, and 20. The two numbers that add up to 5 are (1,4) and (3,2). There are no possibilities that make up 10. The two numbers that add up to 15 are (3,12), (5,10), (7,8), (9,6), (11,4). There are no possibilities to make up 20. So out of 36 total possibilities, there are 7 possibilities for which the sum is divisible by 5. The probability is 7/36.

Dec 16. A jar contains 28 red jelly beans, 14 black jelly beans and 6 green jelly beans. What is the probability that two jelly beans selected at random, and without replacement, from this jar are the same color? Express your answer as a common fraction. (Calculator)
[Solution from Om] The chance of getting two red jbs is 28/48 (because there are 28 red jbs and 48 jbs in total)*27/47 (because there are now only 27 red jbs and 47 jbs in all because you ate one!:) That leads to a 28*27/(48*47) chance of getting two red jbs in a row. You do the same for all the other colored jbs. There is a 28*27/(48*47) chance of getting two red jbs, a 14*13/(48*47) chance of getting two black jbs, and a 6*5/(48*47) chance of getting two green jbs. You add up the three probabilities to get a 968/2256 chance of getting two jbs of the same color. I simplified it to 121/282 using a calculator.

Dec 17. The Chug-A-Long Train Company boxes toy trains with either 5 cars or 7 cars per box. The trains in stock have a total of 53 cars. If Charles selects a box at random, what is the probability that the box contains 7 cars? Express your answer as a common fraction.
[Solution from Gyan] Since the store has a stock of 53 cars, the number of boxes with 5 cars has to be 5 and the number of boxes of 7 cars has to be 4. So the probability is 4 boxes of 7 cars out of 9 boxes in total, which is 4/9.

Dec 18. Jack randomly chooses one of the positive integer divisors of 20, and Jill randomly picks one of the positive integer divisors of 30. What is the probability that Jack and Jill pick the same number? Express your answer as a common fraction.
[Solution from Gyan] There are 6 possible divisors of 20: 1, 2, 4, 5, 10 and 20.  There are 8 possible divisors of 30:  1, 2, 3, 5, 6, 10, 15, and 30. So there is a 1/6 chance of picking 1 out of the divisors of 20 and 1/8 chance of picking 1 out of the divisors of 30. Therefore, the probability that two people both pick 1 is 1/6*1/8 = 1/48. Since the same can be said for the common divisors 2, 5, and 10, the probability is also 1/48. So the probability that the two people both pick the same number is 4*1/48 = 1/12.

Week of Dec 5 – 11.  Let’s start with simple calculation questions. Click GoogleForm to submit answers by Friday night.

Answers: 495, 49, 50, 0, 26, 480, 234

Dec 5. What is the sum of the two-digit multiples of 11?

Dec 6. What is the sum of the prime factors of 2015? (calculator)

Dec 7. What is the average of the first 99 counting numbers?

Dec 8. Koka-Kola is sold in packages of eight 12-ounce bottles. Pepsy-Kola is sold in packages of six 16-ounce bottles. In ounces, what is the absolute difference in the total number of ounces in a package of Koka-Kola and a package of Pepsy-Kola?

Dec 9. At a particular restaurant, hamburgers are priced $3 each, 2 for $5 and 5 for $9. What is the maximum number of hamburgers that can be purchased for $48?

Dec 10. Square tiles with 4-inch sides are 20¢ each, and square tiles with 6-inch sides are 40¢ each. How much will Jerry save tiling a 4-foot by 6-foot floor with the 6-inch tiles laid side by side instead of the 4-inch tiles laid side by side? (answer in cents)

Dec 11. A shirt company charges a one-time setup fee plus a certain price per shirt. An order of 10 shirts costs $84. An order of 20 shirts costs $159. How much does an order of 30 shirts cost?