document.write( "Question 712268: If m and p are positive integers and (m+p) x m is even, which of the following must be true. A)if m is odd, then p is odd B)if m is odd, then p is even C)if m is even, then p is even D)if m is even, then p is odd. \n" ); document.write( "
Algebra.Com's Answer #437845 by jim_thompson5910(35256)![]() ![]() ![]() You can put this solution on YOUR website! (m+p) x m is even \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "so either m is even or m+p is even\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If m is even, then m+p is odd only if p is odd.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If m+p is even, then \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "a) both m and p are even (since even + even = even)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "or\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "b) both m and p are odd (since odd + odd = even)\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "---------------------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So let's go through the choices\r \n" ); document.write( "\n" ); document.write( "A)if m is odd, then p is odd \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "This is true because m+p is even if both m and p are odd and m+p must be even for (m+p) x m to be even (since m is odd)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "--------------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "B)if m is odd, then p is even \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "This is false because it's the complete opposite of choice A which was true.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "--------------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "C)if m is even, then p is even \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If m is even, then (m+p) x m is even (since even x odd = odd x even = even x even = even)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So because p could be even or odd (it wouldn't change the outcome), we can't say for sure if p is even\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So this is false\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "--------------------------------------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "D)if m is even, then p is odd.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Using the same logic above, we can't determine if p is odd when m is even. So there's no way of knowing.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So this is false too. \n" ); document.write( " |