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nick1816
Let n be a positive integer that has the same digit appearing in all of its place values. What is the number of digits in the smallest possible value of n such that it is divisible by all positive digits from 1 to 9, except for 5?

A. 3
B. 6
C. 9
D. 12
E. 18
We are looking for a number with every digit same and which is divisible by 7, 8 and 9

For a number to be divisible by 8, last 3 digits need to be divisible by 8 which makes the minimum number to be 888.

For a number to be divisible by 9, no. of digits or sum of digits should be divisible by 9, in this case it would be number of digits which results into 888888888 (9 digits) or 888888888888888888 (18 digits)

Only one of the above 2 numbers should be divisible by 7 for exactly one choice to be correct, so dividing 888888888 by 7 gives us quotient 126984126 and remainder 6. So it's safe to guess answer would be 18 digits.

PS: If you want to validate choice E, observe the quotient 126984126 when we divided by 7, you would observe there's a pattern which repeats after 6 digits, this is because 111111 is exactly divisible by 7. So any no. of grouping of these 6 digits should always be divisible by 7, like 111111111111 (12 digits) or 111111111111111111 (18 digits).

IMO: E
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