Question

The positive difference between a pair of primes is equal to $2,$ and the positive difference between the cubes of the two primes is $31106.$ What is the sum of the digits of the least prime that is greater than those two primes?

$\ 8$

$\ 10$

$\ 11$

$\ 13$

$\ 16$

Answer: E

Concepts:prime sum and difference of cubes digits

Difficulty rating: 1140

Solution:

Since the primes are $2$ away from each other, we can make them equal to $m-1,m+1,$ where $m$ is their average.

Then, $(m+1)^3-(m-1)^3 = 31106 ,$ making $m^3+3m^2+3m+1$ $-(m^3-3m^2+3m-1)$ $= 6m^2+2 = 31106.$

Therefore, $m^2= 5184,$ so $m=72.$

The primes are therefore $71,73.$ The least prime greater than both of those is $79,$ and its digit sum is $16.$

Thus, the answer is E .

2022 AMC 10B Problem 13

Answer: E

Difficulty rating: 1140

Solution:

Problem 13 in Other Years