document.write( "Question 364536: half of me is greater than 20, one of my digits is double the other ,one of my digits is a perfect square. what number am i ? am i a prime or composite ? \n" ); document.write( "

Algebra.Com's Answer #259942 by solver91311(24713) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "If one of the digits is a perfect square, then that digit has to be 1, 4, or 9 because these are the only perfect square single digits.\r
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\n" ); document.write( "\n" ); document.write( "If the perfect square digit is 1, then the 'double the other digit' has to be 2 (you can't have a digit of 1/2). The largest number you can make from 2 and 1 is 21, but half of 21 is not greater than 20. Therefore, exclude 1.\r
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\n" ); document.write( "\n" ); document.write( "If the perfect square digit is 9, then there is no 'double the other digit' because half of 9 is 4.5 which can't be a digit, and double 9 is 18 which is two digits and therefore can't be a digit. Exclude 9.\r
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\n" ); document.write( "\n" ); document.write( "If the perfect square digit is 4, then half of 4 is 2. You could have 24, but half of 24 is 12 which is not greater than 20. You could have 42 which fits all criteria. But half of 8 is 4, so therefore you could have either 48 or 84 that also fit all three criteria.\r
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\n" ); document.write( "\n" ); document.write( "Hence there are three answers: 42, 48, and 84. All three are obviously composite because they are even numbers greater than 2.\r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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