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Problem 1499

National Olympiad, first round
Number theory Difficulty 6.0 Prove it Philippine Mathematical Olympiad · Philippines

Let NN be a natural number whose base-2016 representation is ABCA B C. Working now in base-10, what is the remainder when N(A+B+C+k)N-(A+B+C+k) is divided by 20152015, if k{1,2,,2015}k \in \{1,2, \ldots, 2015\}?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Solution:

Let NN be the number whose base-20162016 representation is ABCA B C. That is,
N=A×20162+B×2016+C N = A \times 2016^2 + B \times 2016 + C
where A,B,CA, B, C are digits in base 20162016, i.e., 0A,B,C20150 \leq A, B, C \leq 2015.

We are to compute the remainder when N(A+B+C+k)N - (A+B+C+k) is divided by 20152015, for k{1,2,,2015}k \in \{1,2, \ldots, 2015\}.

First, note that 2016=2015+12016 = 2015 + 1.

Let us compute NN modulo 20152015:

20161(mod2015)2016 \equiv 1 \pmod{2015}, so 2016212=1(mod2015)2016^2 \equiv 1^2 = 1 \pmod{2015}.

Therefore,
NA×1+B×1+C(mod2015) N \equiv A \times 1 + B \times 1 + C \pmod{2015}
So NA+B+C(mod2015)N \equiv A + B + C \pmod{2015}.

Therefore,
N(A+B+C+k)(A+B+C)(A+B+C)kk(mod2015) N - (A+B+C+k) \equiv (A+B+C) - (A+B+C) - k \equiv -k \pmod{2015}
So the remainder when N(A+B+C+k)N - (A+B+C+k) is divided by 20152015 is 2015k2015 - k, unless k=2015k = 2015, in which case the remainder is 00.

Final answer: 2015k2015 - k.

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