Answer:
102,232 = 2·2·2·13·983
Explanation:
102,232 ends in 32, so is divisible by 4
102,232/4 = 25558 ends in 2, so is divisible by 2
25558/2 = 12779
is not even, so is not divisible by 2
has a sum of digits mod 9 of 8, so is not divisible by 3 or 9
does not end in 0 or 5, so is not divisible by 5
gives the result 1277-2·9 = 1259; 125-2·9 = 107; 10-2·7 = -4, which is not zero, so 12779 is not divisible by 7
gives the result (1+7+9)-(2+7) = 8, which is not divisible by 11, so 12779 is not divisible by 11
gives the result 1277 +4·9 = 1313, which is divisible by 13, so 12779 is divisible by 13
12779/13 = 983 needs to be tested for divisibility by primes between 17 and 31, namely 17, 19, 23, 29, 31. It is not divisible by any of these, so this is the last prime factor of 102,232:
102,323 = 2·2·2·13·983
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Divisibility rules:
Note that for any transformation rule, the rule can be repeated on the reduced result. The digits of a sum of digits can be summed, for example.
2: ends in even digit
3: sum of digits is divisible by 3
5: ends in 0 or 5
7: subtracting 2 times the last digit from the rest is divisible by 7
11: difference of sums of alternate digits is divisible by 11
13: there are several rules. A couple of useful ones are
- the sum of 4 times the last digit and the remaining number will be divisible by 13 (used above)
- the difference of the sums of alternate groups of 3 digits will be divisible by 13 (232-102 = 130, for the number above; is divisible)
17: subtracting 5 times the last digit from the rest is divisible by 17
19: the sum of 4 times the last 2 digits and the remaining number will be divisible by 19
23: the sum of 3 times the last 2 digits and the remaining number will be divisible by 23
29: the sum of 9 times the last 2 digits and the remaining number will be divisible by 29
31: subtracting 3 times the last digit from the rest is divisible by 31