Tuesday, June 12, 2012

this week's MGRE Math Beast Challenge

From here:

"The Prime of Life"

In a family of four people, none of the people [has] the same age, but all are a prime number of years old. Two of the people are less than 12 years old, and the other two people are between 40 and 52 years old. If the average of their four ages is also a prime number, what are the ages of the family members?

Indicate four such ages (check 4 slots).

( ) 2
( ) 3
( ) 5
( ) 7
( ) 11
( ) 41
( ) 43
( ) 47

Go to it! My own answer will appear in the comments section.



Kevin Kim said...

I was able to brute-force this problem in about a minute. The average of four numbers will be a prime number; the sum of four prime numbers needs to be a number that is evenly divisible by 4. What are the possible primes?

For the first two people:

2, 3, 5, 7, 11 years old

For the second two people:

41, 43, 47 years old

I arbitrarily picked 3, 5, 41, and 43, which seemed to add up to a number evenly divisible by 4, and I believe I got it right on the first try:

3, 5, 41, 43 = all primes

Their sum = 92

92/4 = 23

23, the average, is also prime.

I know that I can't use 2, because the inclusion of 2 in a sum of four numbers (when averaging) would lead to an odd-numbered sum, and such a sum cannot be evenly divided by 4.

Lucky guess on my part, perhaps, but even if I hadn't stumbled upon the correct four numbers right away, I'd have gotten them in the next minute.

My answer, then, is:

3, 5, 41, 43.

Charles said...

Got the same answer by pretty much the same process. I knew I couldn't use 2, and 3/5/41/43 were the first numbers after that in each set.

I was just curious to see if there was some formula that could be used to figure this out, or if plug-and-chug was the only viable method.

Kevin Kim said...

Color me equally curious.