document.write( "Question 784152: I am not grasping this.
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document.write( "Is there any way you can walk me through.
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document.write( " CA\r
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document.write( "Find all possible values for A, B, and C.\r
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document.write( "Each letter represents a single digit from 0-9. Find all the possible values if each digit that make the arithmetic exercise true. \n" );
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Algebra.Com's Answer #477014 by solver91311(24713)![]() ![]() You can put this solution on YOUR website! \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "CA is equal to 4 times AB, since 4 times anything must be even, CA must be even, and therefore A must be even. A cannot be 3 or larger because the sum would then be a three digit number. Hence, A is either 2 or 0. Let A represent 2, then 4 times B is either 2 or 12. If 4 times B is 2, then B would have to be 1/2 which is not a digit 0-9, so if A is 2, 4 times B must be 12, which means B is 3. Then 4 times 2 is 8 plus the carry is 9 and C is 9. And 4 times 23 is indeed 92.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "What is not clear is whether the configuration AB implies that A cannot be zero. I would assume not, however if it can be zero, then you have 4 times B ends in zero, so B must be 5 and then C is 2. And 4 times 5 is 20.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "John \n" ); document.write( " \n" ); document.write( "Egw to Beta kai to Sigma \n" ); document.write( "My calculator said it, I believe it, that settles it \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |