2022 AMC 10B Problem 13

Problem 13 of 25IntermediateAlgebra Number Theory

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?

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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,

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 .

Tagged: prime · sum and difference of cubes · digits