82 lines
4.8 KiB
HTML
82 lines
4.8 KiB
HTML
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<h2>Sum of squares of divisors</h2><div class="info" style="cursor:help;width:200px;margin-bottom:10px;"><h3>Problem 401</h3><span style="width:300px;color:#666;">Published on Saturday, 10th November 2012, 04:00 pm; Solved by 793</span></div>
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The divisors of 6 are 1,2,3 and 6.<BR />
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The sum of the squares of these numbers is 1+4+9+36=50.
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Let sigma2(n) represent the sum of the squares of the divisors of n.
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Thus sigma2(6)=50.
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Let SIGMA2 represent the summatory function of sigma2, that is SIGMA2(n)=<img src='images/symbol_sum.gif' width='11' height='14' alt='∑' border='0' style='vertical-align:middle;' />sigma2(i) for i=1 to n.<BR />
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The first 6 values of SIGMA2 are: 1,6,16,37,63 and 113.
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Find SIGMA2(10<sup>15</sup>) modulo 10<sup>9</sup>.
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