Maths Olympiad Prep

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, 2019

Number theory Difficulty 4.3 AIME Find the answer United States

Problem:

In 2019, a team, including professor Andrew Sutherland of MIT, found three cubes of integers which sum to 4242:
42=(8053873881207597_)3+(80435758145817515)3+(12602123297335631)3 42 = \left(-8053873881207597\_\right)^3 + (80435758145817515)^3 + (12602123297335631)^3
One of the digits, labeled by an underscore, is missing. What is that digit?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

Let the missing digit be xx. Then, taking the equation modulo 1010, we see that 2x3+53+132 \equiv -x^{3} + 5^{3} + 1^{3}. This simplifies to x34(mod10)x^{3} \equiv 4 \pmod{10}, which gives a unique solution of x=4x = 4.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.