document.write( "Question 53140: Can someone help me on this problem?\r
\n" ); document.write( "\n" ); document.write( " I need to list all the factors of 45\r
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Algebra.Com's Answer #35474 by rchill(405) $\"\"$ $\"About$

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To fully factor a number, all the factors must be prime numbers. Primes can only be divided evenly by 1 and themselves. So, some prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23... They go on and on! A trick is that if any number, no matter how many digits are in that number, ends in (or has a last digit of) a 0, 2, 4, 6, or 8, the number can't be prime and can be factored -- because those numbers can always be evenly divided by 2. Another hint is that if any number, again it doesn't matter how many digits long the number is, ends in a 5, then that number isn't prime as it can be evenly divided by 5.\r
\n" ); document.write( "\n" ); document.write( "So, for 45, ending in a 5, let's first divide by 5: $\"45%2F5=9\"$ . So far we have factors of 5 and 9; 5 is prime, but 9 is not, so we continue to factor 9, which factors into 3*3. Because 3 is prime, we have found all our factors of 45... 5*3*3. \n" ); document.write( "

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