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HMMT November, 2013

Algebra Difficulty 4.2 AIME Find the answer United States

Problem:
Find the remainder when 12+32+52+⋯+9921^{2} + 3^{2} + 5^{2} + \cdots + 99^{2} is divided by 10001000.

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

Solution

Solution:
We have S=∑i=049(2i+1)2=∑i=049(4i2+4i+1)=4⋅∑i=049i2+4⋅∑i=049i+50S = \sum_{i=0}^{49} (2i+1)^{2} = \sum_{i=0}^{49} \left(4i^{2} + 4i + 1\right) = 4 \cdot \sum_{i=0}^{49} i^{2} + 4 \cdot \sum_{i=0}^{49} i + 50.

Now,
∑i=049i=49⋅502=1225\sum_{i=0}^{49} i = \frac{49 \cdot 50}{2} = 1225,
∑i=049i2=49⋅50⋅996=40425\sum_{i=0}^{49} i^{2} = \frac{49 \cdot 50 \cdot 99}{6} = 40425.

So,
S=4⋅40425+4⋅1225+50=161700+4900+50=166650S = 4 \cdot 40425 + 4 \cdot 1225 + 50 = 161700 + 4900 + 50 = 166650.

Now, 166650 mod 1000=650166650 \bmod 1000 = 650.

Therefore, the remainder is 650\boxed{650}.

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