Let S be the number of trailing zeroes in ( 2 5 0 0 i + − 2 5 0 0 i ) 2 5 0 0 , then find the sum of all the positive integral divisors of S .
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Awesome solution. Thanks!
A slight shortcut for the last part: σ ( 3 7 5 0 ) = ( 1 + 2 ) ( 1 + 3 ) ( 1 + 5 + 5 2 + 5 3 + 5 4 ) .
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We use that ( 1 + i ) 2 = 2 i . Hence we begin to manipulate the given expression:
( 2 5 0 0 i + − 2 5 0 0 i ) 2 5 0 0 = ( 2 5 0 0 i ( 1 + i ) ) 2 5 0 0 = ( 2 5 0 0 i ( 2 i ) ) 1 2 5 0 = 5 0 0 0 1 2 5 0
Now, it is obvious that there are 1 2 5 0 strings of 0 0 0 trailing, ie there are 3 × 1 2 5 0 = 3 7 5 0 trailing zeroes. We see that 3 7 5 0 = 2 × 3 × 5 4 . Every divisor is of the form 5 n , 2 × 5 n , 3 × 5 n or 6 × 5 n ; these forms are exhaustive and exclusive. Thus,
σ 1 ( 3 7 5 0 ) = n = 0 ∑ 4 5 n + 2 n = 0 ∑ 4 5 n + 3 n = 0 ∑ 4 5 n + 6 n = 0 ∑ 4 5 n = 1 2 n = 0 ∑ 4 5 n = 1 2 × 5 − 1 5 5 − 1 = 9 3 7 2