Problem 111

Number Theory Level 2

$n = \underbrace{1111111\ldots1}_{m \text{ 1's}}$

Find the smallest value of $m>90$ such that $n$ is not a prime number .

The answer is 91.

2 solutions

Zee Ell
Aug 10, 2016

Let's have a look at the smallest value, which is bigger than 90 (91)!

Since 91 = 7 × 13, it is easy to see (just start the division in the conventionial elementary way, but writing the digits of our original number in 13 groups of 7 digits, abbreviating the repeating digits) that a number made out of 91 "1" digits is divisible by 1 111 111 (7 "1" digits).

As a result, we will get 12 groups of "1 000 000" after each other and a 1 in the end (which we will get when dividing the last, 13th group of 7 "1"s).

Therefore, our solution is $\boxed {91}$ .

(And then, we can factorise it further, e.g. 1 111 111 = 239 × 4649 .)

Achal Jain
Aug 11, 2016

if a repunit has prime digits then it is a prime and if composite then it is a composite.

Problem 111

The answer is 91.

2 solutions

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