1. ## repunits

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.

2. 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.

3. Hi Soroban,
Thank you very much. I hope that I can use the hint to solve this question.

My friend came up this :

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

Is it useful regarding to the hint?

4. Originally Posted by Jenny20
Hi Soroban,
Thank you very much. I hope that I can use the hint to solve this question.

My friend came up this :

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

Is it useful regarding to the hint?
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.

5. Hi perfecthacker,

Can you use the " hint " to show me the way to solve my question? If possible, please do show me the working steps. Thank you very much.

6. Originally Posted by Jenny20
Hi perfecthacker,

Can you use the " hint " to show me the way to solve my question? If possible, please do show me the working steps. Thank you very much.
You can combine the hint with soroban's method, as:

17.N=(10^n-1)/9

means:

153.N=10^n-1,

but the rigt hand side is a number all of whoes digits are 9, so you can do
the long division with 153 instead of 17 and a string of 9's instead of 1's
in soroban's method

RonL