document.write( "Question 252666: If a and b are different positive integers, and 5a + 2b = 34 is an integer,
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document.write( "what is the sum of all possible values of a?\r
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document.write( "A) 6 B) 10 C) 12 D) 14 E) 16 \n" );
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Algebra.Com's Answer #184696 by richwmiller(17219)![]() ![]() You can put this solution on YOUR website! 5a+2b=34 \n" ); document.write( "Answers could be a=6 and b=2 \n" ); document.write( "2b must be even but b can be even or odd. \n" ); document.write( "so if 2b must be even then 5a must be even so a must be even since 5 is odd and 34 is even. \n" ); document.write( "If you subtract an even number from an even number, you get an even number. \n" ); document.write( "a must be less than 7 since 5*7=35 and both a and b must be positive. \n" ); document.write( "But the answers are sums of a not a itself \n" ); document.write( "10 is possible because 5*4+2*7=34 \n" ); document.write( "6+4=10\r \n" ); document.write( "\n" ); document.write( "5*2+2*12=34 \n" ); document.write( "so we could have 6,4 and 2 for a total of 12 \n" ); document.write( "since 6, 4 and 2 are possible values for a 14 can't be since it would need 2 twice. \n" ); document.write( "16 has a similar problem with 4 \n" ); document.write( "since we already have a sum of 12 a sum of 16 would need 4 again. \n" ); document.write( "so the highest sum possible for a would be 12 using 2,4 and 6. We could also have a =0 and b would equal 17 but the sum of the a's would not change.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |