document.write( "Question 437675: How many numbers can be obtained as the product of two or more of the numbers 3,4,4,5,5,6,7,7,7? \n" ); document.write( "

Algebra.Com's Answer #302776 by richard1234(7193) $\"\"$ $\"About$

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Note that picking a certain subset from {3,4,4,5,5,6,7,7,7} will uniquely determine the prime factorization of such a number. Even if we pick 6 (which is not coprime to 3 or 4), we will still have a number that is uniquely determined, because the exponent of 2 will be odd regardless of what other numbers are picked.\r
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\n" ); document.write( "\n" ); document.write( "Hence, we have two ways to pick the number of 3's (0 3's or 1 3), three ways to pick the number of 4's, three ways for 5's, two ways for 6's, and four ways for 7's. Multiplying, this becomes 2*3*3*2*4 = 144. However, we need to subtract the size of the set obtained when we pick zero or one numbers. This is easy to count; only {1, 3, 4, 5, 6, 7} can only be picked by choosing zero or one elements. Thus the total number of numbers that satisfy is 144 - 6 = 138.\r
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\n" ); document.write( "\n" ); document.write( "It looks like you had proposed a solution obtaining 212 factors. The error with your solution is that not all of the 212 numbers can be obtained in this way. You let $\"N+=+%282%5E5%29%283%5E2%29%285%5E2%29%287%5E3%29\"$ , which implies that a number such as 8 can be made this way. However, 8 cannot be obtained using the original method. \n" ); document.write( "

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