An algebra problem by Razing Thunder

Algebra Level 4

Find the sum of digits of ( 1 0 4 n 2 + 8 + 1 ) 2 (10^{4n^{2}+8}\ + 1)^{2} , where n n is a positive integer.


The answer is 4.

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2 solutions

Chew-Seong Cheong
Jul 10, 2020

Let m = 4 n 2 + 8 m=4n^2+8 . Then

( 1 0 4 n 2 + 8 + 1 ) 2 = ( 1 0 m + 1 ) 2 = 1 0 2 m + 2 × 1 0 m + 1 = 1 00000 00000 Number of 0s = m 1 2 00000 00000 Number of 0s = m 1 1 \begin{aligned} \left(10^{4n^2+8} + 1 \right)^2 & = \left(10^m + 1 \right)^2 \\ & = 10^{2m} + 2 \times 10^m + 1 \\ & = 1 \underbrace{00000 \cdots 00000}_{\text{Number of 0s}=m-1}2 \underbrace{00000 \cdots 00000}_{\text{Number of 0s}=m-1} 1 \end{aligned}

Therefore the sum of digits is 1 + 2 + 1 = 4 1+2+1 = \boxed 4 .

Brilliant! I have used same way.

Sahil Goyat
Jul 9, 2020

if you use latex your solution must be more impressive

Razing Thunder - 11 months, 1 week ago

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