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2021/22 AHSMC Part 1 Problem 6

Problem 6 of 16

Let AA be a positive integer so that 25303⋅8202=A+125^{303} \cdot 8^{202} = \sqrt{A+1}. The sum of all digits of the integer AA is

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Solution

We have 25303⋅8202=A+1  ⟺  (52)303⋅(23)202=A+1  ⟺  5606⋅2606=A+1  ⟺  10606=A+1  ⟺  (10606)2=A+1  ⟺  101212=A+1  ⟺  A=101212−1,\begin{align*} 25^{303} \cdot 8^{202} = \sqrt{A+1} &\iff (5^2)^{303} \cdot (2^3)^{202} = \sqrt{A+1} \\ &\iff 5^{606} \cdot 2^{606} = \sqrt{A+1} \\ &\iff 10^{606} = \sqrt{A+1} \\ &\iff \bigl(10^{606}\bigr)^2 = A+1 \\ &\iff 10^{1212} = A+1 \\ &\iff A = 10^{1212} - 1, \end{align*} so A=999…9A = 999\ldots 9 where in total there are 12121212 digits of 99. Thus the sum of the digits of AA is equal to 1212×9=109081212 \times 9 = 10908. The answer is (E).

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