Monday, June 18, 2012

answer to last week's MGRE Math Beast Challenge

Yes! I got the answer to last week's MGRE Math Beast Challenge problem correct! The answer was indeed 3, 5, 41, and 43. Here's MGRE's explanation:

The prime numbers less than 12 are 2, 3, 5, 7, and 11. There are five possible ages for the two children, or 5!/2!3! = (5)(4)/(2)(1) = 10 possible combinations for the children’s ages.

The prime numbers between 40 and 52 are 41, 43, and 47. There are three possible ages for the two adults, or 3!/2!1! = 3 possible combinations for the adults’ ages.

In total, there are (10)(3) = 30 possible age combinations—far too many to test each scenario looking for a prime number average age.

An alternative is to start from the resulting average age. The minimum ages of the family members are 2, 3, 41, and 43, which average to 22.25. The maximum ages of the family members are 7, 11, 43, and 47, which average to 27. The only prime number in this range is 23, which implies an age sum of (4)(23) = 92.

If the adults are 43 and 47, the sum of their ages is 90. The sum of the children’s ages would need to be 92 – 90 = 2. The minimum sum of the children’s ages is 2 + 3 = 5, so no need to continue checking these possibilities.

If the adults are 41 and 47, the sum of their ages is 88. The sum of the children’s ages would need to be 92 – 88 = 4. The minimum sum of the children’s ages is 2 + 3 = 5, so again, no need to continue checking these possibilities.

If the adults are 41 and 43, the sum of their ages is 84. The sum of the children’s ages would need to be 92 – 84 = 8. This is only possible if the children are 3 and 5.

Check: (3 + 5 + 41 + 43) = 92, so the average is 92/4 = 23, which is prime.

The correct answers are 3, 5, 41, and 43.

Interesting deductive process! I think I arrived at my answer more intuitively.

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