Maths Olympiad Prep

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, 2013

Number theory Difficulty 4.2 AIME Find the answer United States

Problem:

Find the rightmost non-zero digit of the expansion of (20)(13!)(20)(13!).

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

Solution

Solution:

We can rewrite this as (10×2)(13×12×11×10×9×8×7×6×5×4×3×2×1)=(103)(2×13×12×11×9×8×7×6×4×3)(10 \times 2)(13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) = (10^3)(2 \times 13 \times 12 \times 11 \times 9 \times 8 \times 7 \times 6 \times 4 \times 3). Multiplying together the units digits for the terms not equal to 1010 reveals that the rightmost non-zero digit is 66.

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