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.
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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.
Hello, Jenny!
Quote:
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.
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,Quote:
Originally Posted by Jenny20
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.
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:Quote:
Originally Posted by Jenny20
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