9
$\begingroup$

I have found this statement ,can you help me to prove this.

If an integer is not divisible by 2 or 5, some multiple of that number in decimal notation is a sequence of only a digit.

OBJECTIVE: Now you are given the number and the only allowable digit, you should report the number of digits of such multiple.

For example you have to find a multiple of 3 which contains only 1's. Then the result is 3 because is 111 (3-digit) divisible by 3. Similarly if you are finding some multiple of 7 which contains only 3's then, the result is 6, because 333333 is divisible by 7.

  • 0
    I love how this elementary (but very interesting) question has three totally different proofs alreafy!2012-05-09

3 Answers 3