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Number Theory Level 3

A $50$ -digit number, divisible by $13$ , is as follows

$\large \underbrace{1111\cdots1111}_{\text{Number of 1s}= 25} m \underbrace{1111 \cdots 1111}_{\text{Number of 1s}=24}$

Find the value of $m$ .

7 1 3 0

2 solutions

Mr. India
Mar 29, 2019

$1001=7×11×13$

So, $111111=1001×111$ is also divisible by $13$

This means that $6$ consecutive $1$ are divisible by $13$

So, $24(6×4)$ $1$ 's are also divisible by $13$

So, both left and right of middle $2$ digits are divisible by $13$

This means that middle two digits also must be divisible by $13$ to make the given number divisible by $13$

So, $\overline{1m}$ is divisible by $13$

So, $\boxed{m=3}$

Even if you don't notice that 111111 is divisible by 13, you can argue that 111111111111 (twelve digits) must be divisible by 13. In fact you can write 111111111111 = 1*10^11+...+1, where each power of 10 has a different modular congruence by 13. So the sum of all of this must be multiple of 13.

Costantino Di Bello - 2 years, 2 months ago

Yeah you are right!

Mr. India - 2 years, 2 months ago

" ... where each power of 10 [from $10^0$ to $10^{11}$ ] has a different modular congruence by 13." That was exactly my thinking at first, but it's not true.

13 divides 1001, so $10^3 = 1000 \equiv -1 \ \text{mod} \ 13$ and $10^6 \equiv (-1)^2 \equiv 1 \ \text{mod} \ 13$ . So modulo 13, the powers of 10 only cycle between six values (not all twelve non-zero values): 10, 9, 12, 3, 4, 1. But of course, these values do sum up to be zero mod 13.

Matthew Feig - 2 years, 2 months ago

You're right, actually I knew this but I was looking for a method to avoid calculations. The idea that 11...1 (p digits) is a multiple of p, with p prime, is supported by the Little Fermat theorem. This is a particular case in which 10^((13-1)/2) is 1 modulus 13.

Costantino Di Bello - 2 years, 2 months ago

Kyle T
Apr 1, 2019

just confirming via wolfram

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