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OSU-Marion Math Challenge 2008-2009
Correct solutions by current OSU-Marion and Delaware Center students will be recognized on this page and in Buckeye Briefs. The student with the most correct solutions at the end of the year will receive the coveted "Golden Calculator" and a prize ($100 Gift Certificate) at the Academic Recognition Program in the Spring.
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Math Challenge Problem # 11
First, find a six-digit number in which the first digit is
four more than the second;the third is one less than the second;
the product of the first two digits equals the two-digit number
formed by reading the third and fourth digits together; the
fifth is the sum of the first and third digits; and the last digit
is the sum of the fourth and fifth digits.
Second, using each of the digits 1 to 9 exactly once, form a 9-digit integer with the property that its first digit is divisible by 1, the number formed from its first two digits is divisible by 2, the number formed from its first three digits is divisible by 3, and so on. Find all such numbers. For example, the number 123459876 is a candidate but fails to be a solution since 1234 is not divisible by 4.
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Due February 22, 2009 |
Solutions by:
Chris Welker |
Math Challenge Problem # 10
The figure below is a map of several city blocks. Determine the number of paths that go from Point A to Point B travelling along the streets only going North or East. Second, determine the number of such paths that avoid the intersection at Point C.

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Due Monday January 26, 2009 |
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Math Challenge Problem #9
Here are a few probability questions. Suppose you are rolling a pair of dice, what is the probability of rolling doubles three times in a row? Now, suppose you roll a die twice, what is the probability that the value of the second throw is more than the first? Third, suppose you roll five dice at once, find the probability of getting a “full house” (i.e. three of one number and to of a different number). Finally, which is more likely to roll a single die six times without getting a 6, or to roll a pair of dice 24 times without getting double 6’s? |
Due Tuesday January 20, 2009 |
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Math Challenge Problem #8
Here are a couple questions about decimal representations of fractions. If p/q is a fraction (in lowest terms) then the decimal representation of the fraction p/q must eventually terminate or start repeating. Can you determine what properties about p and q make a fraction that terminates? If the decimal representation repeats, the repeating sequence might not start right at the beginning (i.e. 679/5500=0.123454545….). Can you determine properties about p and q that determine how many decimal places occur before the repeating pattern begins and how many decimal places are in the repeating pattern?
Can you find fractions (in lowest terms) that equal 0.145314531453….., 0.9871111111…
and 0.200812121212….? |
Due Monday, December 1, 2008 |
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Math Challenge Problem #7
A large water tank has two inlet pipes (a large one and a small one) and one outlet pipe. It takes 2 hours to fill the tank with the large inlet pipe. On the other hand, it takes 5 hours to fill the tank with the small inlet pipe. The outlet pipe allows the full tank to be emptied in 7 hours.
What fraction of the tank (initially empty) will be filled in 90 minutes if all three pipes are in operation? When will the tank be filled?
Solve the two questions again assuming that the small inlet pipe was only turned on after the others were operating for half an hour. |
Due Monday, November 24, 2008 |
Solutions By:
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Math Challenge Problem #6
Goldilocks, the chemist, is trying to mix a container of 70% acid solution. She starts with 10 liters of pure HCL acid. She adds a certain amount of pure water to dilute it. Next, she drains off 1/3 of the mixture and replaces it with more pure acid. Now it is too strong. So she adds 5 liters of pure water. Now it is too weak. Finally, she drains off one fourth of the mixture and replaces it with pure acid. Now it is just right at 70% solution. Your question is how much water did she add at the start?
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Due Monday, November 10, 2008 |
Solutions by:
Chris Welker
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Math Challenge Problem #5
The Math club has a unique way of choosing the club president from the three candidates. You are competing against Alice and Bob for the honor. The three of you are shown five hats, three of which are red and two are blue. Then you are blindfolded and placed in a line, all facing forward with Alice in back, Bob in the middle and you in the front. One of the five hats is placed on each of you. The blindfolds are removed so that now Alice can see both you and Bob, while Bob can only see you. You can’t see either of them. The first person to determine the color of the hat on their head becomes president. After a minute or so, nobody has spoken up. Could you become president? Explain how you would determine the color of your hat. |
Due Monday, November 3, 2008 |
Solutions by:
Chris Welker
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Math Challenge Problem #4
Here are some ‘24’ puzzles. Use the four numbers given to add/subtract/multiply and/or divide to get a final answer of 24. For example, if you are given the numbers {1,2,6,7}, you can get 24 as follows: (7-1)*(6-2) = 24. Or use the numbers {2,2,6,6} like this: (6/2)*(6+2) = 24. There’s often more than one way to solve it.
Now it’s your turn. Here are five sets of numbers (and one bonus problem) First, use the numbers {3,4,5,8} to get 24. Second, use {2,6,9,9}. Third, use {1,2,5,8}. Fourth, use {4,5,6,8}. Finally, use {3,3,5,9}. Bonus problem, use {3,3,8,8} to get 24 (I think this one is pretty hard). |
Due Monday October 27, 2008
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Solutions by:
Brittany Hamilton
Nick Leeper
Mary Mooney
Edam Mossbrugger
Hallie Prenslow
Renee Rush
Chris Welker
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Math Challenge Problem #3
Group the consecutive counting numbers as follows: (1), (2, 3), (4, 5, 6), (7, 8, 9, 10), (11,12,13,14,15), . . . Note there is one number in the first group, two in the second group, three in the third, etc. Determine the sum of the 19 numbers in the 19th group? How about the 199 numbers in the 199th group? Explain your reasoning. Please, don’t just right out the first 199 groups…
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Due Monday October 20, 2008 |
Solutions by:
Chris Welker
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Math Challenge Problem #2
Suppose there are three boxes in front of you. You are told that each box contains two balls. One box has two white balls, another one has two black balls and the third one has a white ball and a black ball, but you don't know which box has what inside. Each box has a label. One label says "TWO WHITE", another label says "TWO BLACK", and the other label says "ONE WHITE & ONE BLACK". However, you also know that all three boxes are labeled incorrectly. Your task is to correctly identify the boxes. You are only allowed to look at only one ball from only one of the three boxes. Which box should you open in order to sort out the labels? Explain your reasoning. |
Due Monday October 13, 2008 |
Solutions by:
Ambrosia Gamble
Britanny Hamilton
Linda Irvine
Hallie Prenslow
Lynda Ross
Chris Welker
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Math Challenge Problem #1
The four corners of a rectangle lie on the coordinate plane at the coordinates (1,2), (5,2), (1,8), and (5,8).
If a point is randomly selected within the rectangle, what is the probability that the y-coordinate of the point is greater than the x-coordinate? |
Due Monday October 6, 2008 |
Solutions by:
Jeff Frazier
Chris Welker
Ambrosia Gamble
Hallie Prenslow
Seth Franke
Nick Leeper |
Math Challenge Archive
2007-2008
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