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Number Theory Level 3

Find the last 2 digits of:

$3^2 + 7^2 + 11^2 + \cdots + 2007^2 .$

The answer is 58.

1 solution

Alex Spagnoletti
May 15, 2016

First we can notice that all the numbers are of the form $4k-1$ , so we can rewrite everything as $\sum_{k=1}^{502} (4k-1)^2$ so it is equal to $\sum_{k=1}^{502} (16k^2-8k+1)$ and applying the formulas for sum of consecutive numbers and consecutive squares we obtain $\dfrac{16*502*503*1005}{6} - \dfrac{8*502*503}{2} + 502$ and working $\pmod {100}$ it is easy to compute $(5*16)-(4*2*3)+2$ which is equal to $58$

Moderator note:

Simple standard approach.

There's a simplest way by seeing that the sum of all these squares up to 1999 give a multiple of 100. So 2003^2+2007^2=3^2+7^2=58

Alessandro Bruno - 5 years ago

Yes, it is true.

Alex Spagnoletti - 5 years ago

Back to 2007

The answer is 58.

1 solution

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