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April 16, 2014

April 16, 2014

Posted by **Jen** on Wednesday, December 23, 2009 at 9:21pm.

- Math -
**MathMate**, Thursday, December 24, 2009 at 8:27amYou may or may not have read about the divisibility test of large numbers for 7 and 13, as follows:

If the number has more than 3 digits, subdivide the number into groups of three digit numbers. Sum the odd groups of numbers, and subtract from the sum of the even groups. If the difference is divisible by 7, the original number also. Likewise for 13.

For example:

Test divisibility of 184,751,749 by 7.

Odd groups - even groups = 184+749-751 = 182

182 is divisible by 7, so is the original number.

182 is divisible by 13, so is the original number.

Back to the given question:

the number sought is 9753xy where xy

are two digits to be found.

As a guess, we put xy=00, so the number to be tested will be 975,300

If 975,300 is divisible by 7 and 13, then so does 975-300=675.

For a number to be divisible by both 7 and 13, it must be divisible by 91.

Our task is therefore reduced to finding the greatest number less than 675 that is divisible by 91. By division, we get 675=91*7+38=637+38

Therefore, if the original number was 975,338, it would be divisible by both 7 and 13.

975-338=637

637/7=91

637/13=7

Use long division of 975,338 to check the divisibility by 7 and 13.

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