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Question from Madison, a student:
what are the factors of 878

Madison,

Find prime factors, then divide them out and try again with the quotient. If a number has no prime factor smaller than its square root it is prime.

You should know that the first few primes are
2,3,5,7,11,13,17,19,23,29...

For a few of these there are easy tests: a number is divisible by

2 if its last digit is even
3 if the sum of its digits is divisible by 3 [repeat test is necessary]
5 if its last digit ends in 0 or 5

Otherwise you need to use long division; for testing you may find it easier to keep only remainders.

Once you have a list of prime factors expressed in the form "pi occurs ni times", systematically list every number with 0,1,...,n copies of p for each p.

Here's an example. Start with 680
680 is bigger than 2 squared SO TEST FOR 2
last digit even, divisible by 2
2 x 340
340 is bigger than 2 squared SO TEST FOR 2
last digit even, divisible by 2
2x2 x 170
170 is bigger than 2 squared SO TEST FOR 2
last digit even, divisible by 2
2x2x2 x 85
85 is bigger than 2 squared SO TEST FOR 2
last digit odd, not divisible by 2, NEXT PRIME
85 is bigger than 3 squared SO TEST FOR 3
add digits: not divisible by 3, NEXT PRIME
85 is bigger than 5 squared SO TEST FOR 5
ends in 5
2x2x2 x 5 x 17
17 is less than 5 squared, LAST PRIME FACTOR

2x2x2 x 5 x 17

To make up a complete list of factors, include every number with:

0,1,2 or 3 2's
0 or 1 5
0 or 1 17

multiplied together (4x2x2 possibilities)

1, 2, 4, 8 5, 10, 20, 40
17,34,68,136 85,170,340,680

is the complete list.

Now try this yourself for 878. (You will have to try some larger primes, all in the list above, that can only be tested by long division. Your final list of factors should be quite short.)

Good hunting!
RD

 

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