Question
Given that N= abcdefghij is a ten-digit number . All of the digits (a, b, c, d, ...) are different from one another. If 1 1 1 1 1 divides it evenly, how many different possibilities are there for abcdefghij?
Answer: Option B
:
B
If all the digits are different, and there are 10 of them, then all the digits from 0 to 9 must appear.
That implies that the digits of abcdefghij add up to 45 =0+1+2+3+...+9.
That implies that abcdefgh is definitely divisible by 9.
Since 9 and 1 1 1 1 1 are relatively prime, their least common multiple is 9×11111=99999,
Thus abcdefgh must be divisible by 99999.
We have to find the condition such that abcdefghij is divisible by 99999.
abcdefghij =abcde×105+ fghij and, this number is exactly divisible by (105 -1).
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:
B
If all the digits are different, and there are 10 of them, then all the digits from 0 to 9 must appear.
That implies that the digits of abcdefghij add up to 45 =0+1+2+3+...+9.
That implies that abcdefgh is definitely divisible by 9.
Since 9 and 1 1 1 1 1 are relatively prime, their least common multiple is 9×11111=99999,
Thus abcdefgh must be divisible by 99999.
We have to find the condition such that abcdefghij is divisible by 99999.
abcdefghij =abcde×105+ fghij and, this number is exactly divisible by (105 -1).
Hence by remainder theorem, (abcde×105 + fghij) must be 99999.
Then the following equations must all hold:
f =9-a, g =9-b, h =9-c, I =9-d, j =9-c
Now there are 9 Options for a (it can't be 0) and then f is known.That leaves 8 options for b, and
then'g' is known. That leaves 6 Options for c, and then h is known.
That leaves 4 Options for d and then I is known.
That leaves 2 Options for c, and then j is known.
Thus, the total number of such number abcdefghij is 9×8×6×4×2=3456.
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