Problem of the Week Archive
Good Luck Mathletes! – February 3, 2014
Problems & Solutions
This past weekend the regional level of the MATHCOUNTS Competition Series was in full swing as
Mathletes® in 36 states competed in the 2014 Chapter Competition. See if you can solve these
problems from the 2013 Chapter Competition.
Zeno’s pet rabbit always jumps halfway to a carrot no matter how far away he is from the carrot. He
will eat the carrot if he lands within 6 inches of it. How many times must Zeno’s rabbit jump to eat a
carrot that is initially 12 feet away?
[Sprint Round #9]
After the first jump, the rabbit’s distance from the carrot is 6 feet. After the second jump, that distance is 3 feet. After the third
jump, the rabbit’s distance from the carrot is 1.5 feet, or 18 inches. After the fourth jump, that distance is 9 inches. After the fifth
jump, the rabbit’s distance from the carrot is 4.5 inches, which means the rabbit has a rather tasty carrot after jumping 5 times.
Shaina has one stick of length a cm and another of length b cm, where a  b. She needs a third
stick with length strictly between 8 cm and 26 cm to make the third side of a triangle. What is the
product ab?
[Team Round #7]
The length of the third side of a triangle is always less than the sum of the lengths of the other two sides, a + b. The length of
the third side must also be greater than the absolute difference of the lengths of the two sides, a – b, where a > b. Therefore,
a + b = 26 and a – b = 8. Solving this system by adding the two equations, we get a + b + a – b = 26 + 8 → 2a = 34 →
a = 17. That means 17 + b = 26 → b = 26 – 17 → b = 9. Therefore, ab = 17 × 9 = 153.
What is the mean of all possible positive three-digit integers in which no digit is repeated and all
digits are prime? Express your answer as a decimal to the nearest hundredth.
[Target Round #4]
The three digits in each number must come from the set of single-digit prime numbers {2, 3, 5, 7}. There are 4C3 = 4!/(3! 1!) =
4 combinations of three numbers chosen from the set of four prime numbers: {2,3,5}, {2, 3, 7}, {2, 5, 7} and {3, 5, 7}. There are
3! = 6 permutations of each of these three-digit combinations for a total of 4 × 6 = 24 three-digit integers. They are 235, 253,
325, 352, 523, 532, 237, 273, 327, 372, 723, 732, 257, 275, 527, 572, 725, 752, 357, 375, 537, 573, 735 and 753. If
we were to add up all the integers, the units column would contain six 2s, six 3s, six 5s and six 7s. That’s (2 × 6) + (3 × 6) +
(5 × 6) + (7 × 6) = 12 + 18 + 30 + 42 = 102. That gives us a 2 in the units column, and we carry 10 over to the tens
column. In the tens column you also have six of each digit. So that’s a sum of 102 plus the 10 we carried over, or 112. That
give us a 2 in the tens column and we carry over 11 to the hundreds column. Similarly, there are six of each digit in the
hundreds column. That’s a sum of 102 plus the 11 we carried, or 113. So the sum of these 24 integers is 11,322. The mean is
11,322/24 = 471.75.
The intersection of a circular region of radius 3 inches and a circular region of radius 2 inches has
area π in2. In square inches, what is the area of the total region covered by the two circular regions?
Express your answer in terms of π.
[Countdown Round #8]
The area of a circle of radius 3 inches is 32π = 9π in2 and the area of a circle of radius 2 inches is 22π = 4π in2. The area of the
region covered by these two circular regions is the sum of the areas of the circles less the area of the overlapping region, π in2.
That’s 9π + 4π – π = 12π in2.