document.write( "Question 793660: How do you find the product of primes for the number 10080? \n" ); document.write( "
Algebra.Com's Answer #480417 by KMST(5328)\"\" \"About 
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To find the prime factorization for any number you look for prime factors.
\n" ); document.write( "First, is the number divisible by 2?
\n" ); document.write( "If so, divide it by 2, and keep on working with the result.
\n" ); document.write( "If not, try the same with 3; then with 5, 7, 11, 13. 17, 19, 23, and all the other prime numbers, until you get to one whose square exceeds the number.
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\n" ); document.write( "10080 divides by 2 evenly and the result is 5040, so \"10080=2%2A5040\"
\n" ); document.write( "5040 divides by 2 evenly and the result is 2520, so \"10080=2%2A5040=2%2A2%2A2520\"
\n" ); document.write( "2520 divides by 2 evenly and the result is 1260, so \"10080=2%2A5040=2%2A2%2A2520=2%2A2%2A2%2A1260\"
\n" ); document.write( "1260 divides by 2 evenly and the result is 630, so \"10080=2%2A5040=2%2A2%2A2520=2%2A2%2A2%2A1260=2%2A2%2A2%2A2%2A630\"
\n" ); document.write( "We really do not have to keep writing the products, we just need to keep track of how many times we divide by 2.
\n" ); document.write( "630 divides by 2 evenly and the result is 315.
\n" ); document.write( "315 is not divisible by 2, but divides evenly by 3 and the result is 105.
\n" ); document.write( "105 divides evenly by 3 and the result is 35.
\n" ); document.write( "35 is not divisible by 3, but divides evenly by 5 and the result is 7.
\n" ); document.write( "So, all the factors that we peeled from 10080, like layers from an onion, are
\n" ); document.write( "\"10080=2%2A2%2A2%2A2%2A2%2A3%2A3%2A5%2A7=2%5E5%2A3%5E3%2A5%2A7\"
\n" ); document.write( "To check we can calculate back the product:
\n" ); document.write( "\"2%5E5=32\" and \"3%5E2=9\", and \"32%2A9%2A5%2A7=10080\" according to my calculator.
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\n" ); document.write( "Of course, if no one asks me to show my work, I take shortcuts, peeling larger numbers in easy chunks. For example I would divide by \"10=2%2A5\" in my head, to get 1008, and would look for factors of 1008, while keeping track of the 2 and 5 (I would write them on scrap paper).
\n" ); document.write( "I may realize that 1008 divides by \"9=3%2A3\", and dividing by 9 would get 112, keeping track of the two 3's in \"9=3%2A3\". Then I would work with 112 to find the remaining factors. Dividing by \"4=2%2A2\" I would get \"28=4%2A7=2%2A2%2A7\".
\n" ); document.write( "So the factor found were 2 and 5 first (as factors of 10),
\n" ); document.write( "3 and 3 next (in 9),
\n" ); document.write( "2 and 2 next (in 4),
\n" ); document.write( "and finally 2, 2, and 7 (in 28).
\n" ); document.write( "In all, 2 appeared 5 times, 3 appeared twice, 5 once, and 7 once,
\n" ); document.write( "so \"10080=2%5E5%2A3%5E2%2A5%2A7\"
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