Find the smallest multiple of 17 that has only 1s in its decimal expansion.

Hint: the numbers in question are of the form (10^n-1)/9.

How should I solve this question? Please teach me . Thank you.

Results 1 to 6 of 6

- December 25th 2006, 05:25 AM #1

- Joined
- Nov 2006
- Posts
- 123

- December 25th 2006, 06:36 AM #2

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 11,826
- Thanks
- 714

Hello, Jenny!

Find the smallest multiple of 17 that has only 1's in its decimal expansion.

Hint: the numbers in question are of the form (10^n - 1)/9

I haven't found a way to use that hint.

If you're desperate, here's a**very**primitive method.

Divide 17 into a "string of 1's" ... and wait for it to*come out even*.Code:6 5 3 5 9 4 7 7 1 2 4 1 8 3 --------------------------------- 1 7 ) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -------------------------------- 0

Therefore: 1,111,111,111,111,111 is the smallest multiple of 17.

- December 25th 2006, 07:29 AM #3

- Joined
- Nov 2006
- Posts
- 123

- December 25th 2006, 07:58 AM #4

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 9

I think you mean,

17m=(10^n-1)/9

The number on left is multiple of 17.

The number on right is a*repunit*.

Multiply by 9,

153m=10^n-1

Thus,

153m+1=10^n

Note, the expression on right is 1 followed by zeros.

Thus, find the smallest "m" that will make,

153m+1

Be followed by zeros.

- December 25th 2006, 08:10 AM #5

- Joined
- Nov 2006
- Posts
- 123

- December 25th 2006, 08:51 AM #6

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 4