Here is a way to proceed.
The sum of the digits of both numbers is a multiple of 3 so both numbers are divisible by 3. Divide them both by 3
\frac{65569}{83657}
Clearly neither divisible by 5
break this number up like this 65569 = 065 569 , into groups of 3 digits each , now subtract, 569 - 65 = 504
This gives us at once a divisibility test for 7, 11, and 13
if 504 is divisible by 7, so is 65569
if 504 is divisible by 11 so is 65569
if 504 is divisible by 13 so is 65569
we see 7 divides 504 but 11 and 13 do not
Do the same for 83657 = 083 657 , subtract 657 - 83 = 574
It's easy to see 574 is divisible by 7 since previously 504 was divisible by 7 and this is 70 more than that. No need to test for 11 and 13 since those factors won't cancel if they are there (they are not)
Divide them both by 7.
\frac{9367}{11951}
perform previous procedure again
367 - 9 = 358 not divisible by 7.
Note* Divisibility test for 3 did not have to be done TWICE because sum of digits was NOT 9 but in general you have to repeat the test until divisibility posibility is exhausted.
For divisibility by 17 , cut off the last digit of 9367 , multiply it by 5 , then subtract that from the truncated number.
936 - 35 = 901
If 901 is divisible by 17 then so is 9367 , it is.
Do the same for 11951,
1195 - 5 = 1190
If 1190 is divisible by 17 , then so is 11951 , it is.
Divide them both by 17
\frac{551}{703}
apply divisibility test again for 17
55 - 5 = 50
not divisible by 17 so neither is 551
For divisibility by 19 , cut off the last digit of 551 , multiply it by 2, add that to the truncated number. 55 + 2 = 57
If 57 is ... well you get the idea , it is
Do the same for 703
70 + 6 = 76 is divisible by 19
Divide them both by 19
\frac{29}{37}
It's quite obvious now you can stop.
:)
