Maths Olympiad Prep

Library / /383 of 1394

, 2016

Algebra Difficulty 5.0 AIME, harder Prove it United States

Problem:
Let AA denote the set of all integers nn such that 1n100001 \leq n \leq 10000, and moreover the sum of the decimal digits of nn is 22. Find the sum of the squares of the elements of AA.

Solution

Solution:
From the given conditions, we want to calculate
i=03j=i3(10i+10j)2 \sum_{i=0}^{3} \sum_{j=i}^{3}\left(10^{i}+10^{j}\right)^{2}
By observing the formula, we notice that each term is an exponent of 1010. 10610^{6} shows up 77 times, 10510^{5} shows up 22 times, 10410^{4} shows up 99 times, 10310^{3} shows up 44 times, 10210^{2} shows up 99 times, 1010 shows 22 times, 11 shows up 77 times. Thus the answer is 72949277294927.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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