Math
posted by Jen .
The first four digits of a 6digit number are 9, 7, 5, 3, in this order. What are its last two digits if the number is divisible by 7 and 13?

You 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+749751 = 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 975300=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.
975338=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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