2500 Followers Problem

Algebra Level 5

Let S S be the number of trailing zeroes in ( 2500 i + 2500 i ) 2500 \left( \sqrt{2500i} +\sqrt{-2500i} \right)^{2500} , then find the sum of all the positive integral divisors of S S .

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The answer is 9372.

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1 solution

Jake Lai
Sep 29, 2015

We use that ( 1 + i ) 2 = 2 i (1+i)^2 = 2i . Hence we begin to manipulate the given expression:

( 2500 i + 2500 i ) 2500 = ( 2500 i ( 1 + i ) ) 2500 = ( 2500 i ( 2 i ) ) 1250 = 500 0 1250 (\sqrt{2500i} + \sqrt{-2500i})^{2500} = (\sqrt{2500i}(1+i))^{2500} = (2500i(2i))^{1250} = 5000^{1250}

Now, it is obvious that there are 1250 1250 strings of 000 000 trailing, ie there are 3 × 1250 = 3750 3 \times 1250 = 3750 trailing zeroes. We see that 3750 = 2 × 3 × 5 4 3750 = 2 \times 3 \times 5^4 . Every divisor is of the form 5 n , 2 × 5 n , 3 × 5 n 5^n, 2 \times 5^n, 3 \times 5^n or 6 × 5 n 6 \times 5^n ; these forms are exhaustive and exclusive. Thus,

σ 1 ( 3750 ) = n = 0 4 5 n + 2 n = 0 4 5 n + 3 n = 0 4 5 n + 6 n = 0 4 5 n = 12 n = 0 4 5 n = 12 × 5 5 1 5 1 = 9372 \sigma_1(3750) = \sum_{n=0}^4 5^n + 2 \sum_{n=0}^4 5^n + 3 \sum_{n=0}^4 5^n + 6 \sum_{n=0}^4 5^n = 12 \sum_{n=0}^4 5^n = 12 \times \frac{5^5-1}{5-1} = \boxed{9372}

Awesome solution. Thanks!

Sandeep Bhardwaj - 5 years, 8 months ago

A slight shortcut for the last part: σ ( 3750 ) = ( 1 + 2 ) ( 1 + 3 ) ( 1 + 5 + 5 2 + 5 3 + 5 4 ) \sigma(3750) = (1+2)(1+3)(1+5+5^2+5^3+5^4) .

Arjen Vreugdenhil - 5 years, 8 months ago

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