MathDB

2013-digit numbers wanted. sum of pairwise product of digits = 1 mod 4

Source: 2013 Argentina OMA Finals L3 p3

January 15, 2023

Digitsnumber theoryremainder

Problem Statement

Find how many are the numbers of

2013

digits

d_1d_2…d_{2013}

with odd digits

d_1,d_2,…,d_{2013}

such that the sum of

1809

terms

d_1 \cdot d_2+d_2\cdot d_3+…+d_{1809}\cdot d_{1810}

has remainder

1

when divided by

4

and the sum of

203

terms

d_{1810}\cdot d_{1811}+d_{1811}\cdot d_{1812}+…+d_{2012}\cdot d_{2013}

has remainder

1

when dividing by

4