SOLUTION: any number of the form "abcabc" must be divisible by which of the following: (1)8 (2)7,13,11 (3)9

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Question 1047093: any number of the form "abcabc" must be divisible by which of the following:
(1)8
(2)7,13,11
(3)9
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
the answer is selection 2.
it must be divisible by 7 and 13 and 11.
i have no idea why.

it must be means there are no exceptions.

i took 123123 and divided it by 8 and 9 and did not get integers, so it can't be 8 or 9.

i took 123123 and divided it by 7 and got an integer.
i then divided it by 13 and got an integer.
i then divided it by 11 and got an integer.

i then took another random number and did the same.

i took 987987
same result.
it was not divisible by 8 or 9.
it was divisible by 7, 13, and 11.

i then took 7 * 13 * 11 and got 1001.

i then divided 123123 by 1001 and got 123

i then divided 987987 by 1001 and got 987.

interesting.........

since 1001 is equal to 1000 + 1, i then looked at:

123 * 1001 = 123 * (1000 + 1) = 123000 + 123 = 123123.

i then looked at:

987 * 1001 = 987 * (1000 + 1) = 987000 + 987 = 987987.

since abc * 1001 is equal to abc * 1000 + abc * 1 which is equal to abcabc, then:

abcabc / 1001 is equal to abc.

since 1001 is equal to (7 * 13 * 11), then:

abc * 1001 is equal to abc * 7 * 13 * 11 which is then equal to abcabc.

therefore abcabc / 1001 is equivalent to abcabc / (7 * 13 * 11).

since it's divisible by 1001, it has to be divisible by the factors of 1001, namely 7, 13, and 11.

that's what i think is happening.





Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.
any number of the form "abcabc" must be divisible by which of the following:
(1) 8
(2) 7,13,11
(3) 9
~~~~~~~~~~~~~~~~~~~ 

1.  Any number of the form "abcabc" is equal to 1001*N, where N = "abc", a 3-digit number written with the digits "a", "b" and "c".


2.  The number 1001 is divisible by 7, 11 and 13.
    ( Actually, 1001 = 7*11*13. )


3.  Therefore, any number of the form "abcabc" is divisible by 7, 11 and 13.

    The answer is: Option 2).



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