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Number Theory Level 2

$A=\underbrace{111\dots111}_{\text{Number of 1s = 81}}$

What is the remainder when $A$ is divided by 81?

9 36 0 54 27 18 72 45

2 solutions

Chew-Seong Cheong
Jul 9, 2017

Let $a = \underbrace{111 111 111}_{\text{\# of 1s = 9}}$ . Then $A=a(1+10^9+10^{18}+ 10^{27}+…+10^{72})$ . Note that the sum of digit of $a$ is 9; $a$ is divisible by 9. And that the sum of digit of $1+10^9+10^{18}+ 10^{27}+…+10^{72}$ is also 9 and it is divisible by 9. Therefore $A$ is divisible by 81 and the remainder is $\boxed{0}$ .

Since you know that the answer is 0, it's actually simpler to prove that $10^{81} -1$ is divisible by 729 (which is equivalent to proving that $A$ is divisible by $729/9 = 81$ ).

Just use binomial expansion:

$10^{81} - 1 = (9+ 1)^{81 } - 1 = 9^3 (\cdots) + (\text{last few terms})$

What's left to do is to prove that the $\text{last few terms}$ is divisible by 729 too.

Pi Han Goh - 3 years, 10 months ago

Ankit Dash
Aug 14, 2017

81 can be written as 3^4. So 'A' should be divisible by 3.Since there are all one's therefore sum of all digits =81 which is divisible by 3.Since 'A' satisfies the divisiblity criteria of 3,hence the remainder is 0

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