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2022 AMC 10B Problem 13

Problem 13 of 25IntermediateAlgebraNumber Theory

The positive difference between a pair of primes is equal to 2,2, and the positive difference between the cubes of the two primes is 31106.31106. 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 22 away from each other, we can make them equal to m1,m+1,m-1,m+1, where mm is their average. Then, (m+1)3(m1)3=31106,(m+1)^3-(m-1)^3 = 31106 , making m3+3m2+3m+1m^3+3m^2+3m+1(m33m2+3m1)-(m^3-3m^2+3m-1) =6m2+2=31106.= 6m^2+2 = 31106. Therefore, m2=5184,m^2= 5184, so m=72.m=72. The primes are therefore 71,73.71,73. The least prime greater than both of those is 79,79, and its digit sum is 16.16. Thus, the answer is E .

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Concepts: prime · sum and difference of cubes · digits

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.