document.write( "Question 242631: If a number is divisible by both 4 and 9, by what other numbers is it divisible?Explain. Thank you. \n" ); document.write( "
Algebra.Com's Answer #177610 by Theo(13342)\"\" \"About 
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since 9 is not a multiple of 4, the smallest number that can be divisible by 4 and 9 has to be a number that is the result of a product of 4 * 9.\r
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\n" ); document.write( "\n" ); document.write( "if it is divisible by 4 and 9, then it has to be divisible by factors of 4 and 9.\r
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\n" ); document.write( "\n" ); document.write( "that means it is divisible by 2 and 3 as well.\r
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\n" ); document.write( "\n" ); document.write( "it should also be divisible by any combination of the factors of 4 and 9.\r
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\n" ); document.write( "\n" ); document.write( "since the factors would be 2*2*3*3, then:\r
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\n" ); document.write( "\n" ); document.write( "that would be:\r
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\n" ); document.write( "\n" ); document.write( "2*2 = 4
\n" ); document.write( "2*3 = 6
\n" ); document.write( "3*3 = 9
\n" ); document.write( "2*2*3 = 12
\n" ); document.write( "2*3*3 = 18\r
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\n" ); document.write( "\n" ); document.write( "4*9 * anything else would yield even more possibilities/.\r
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\n" ); document.write( "\n" ); document.write( "example:\r
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\n" ); document.write( "\n" ); document.write( "4*9*2 = 72\r
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\n" ); document.write( "\n" ); document.write( "since there's another factor in the mix, 72 should be divisible by even more numbers that are combinations of factors of 2*2*3*3*2\r
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\n" ); document.write( "\n" ); document.write( "those would be:\r
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\n" ); document.write( "\n" ); document.write( "2*2 = 4
\n" ); document.write( "2*3 = 6
\n" ); document.write( "3*3 = 9
\n" ); document.write( "2*2*3 = 12
\n" ); document.write( "2*3*3 = 18
\n" ); document.write( "plus:
\n" ); document.write( "2*2*2 = 8
\n" ); document.write( "2*2*3*2 = 24
\n" ); document.write( "2*3*3*2 = 36\r
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