document.write( "Question 995336: In the correct addition shown at the right, A,B, and C are different non-zero digits. What is the value of C?
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document.write( " BB
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document.write( "+BB
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document.write( "-------
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document.write( "ABC
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document.write( "
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document.write( "A)0
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document.write( "B)6
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document.write( "C)8
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document.write( "D)9 \n" );
document.write( "
Algebra.Com's Answer #614088 by jim_thompson5910(35256)![]() ![]() ![]() You can put this solution on YOUR website! Let's go through all the choices\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "A) 0\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If C = 0, then B+B ends with a zero and this is only possible if B = 0 or B = 5. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "if B = 0, then A,B,C all must be 0 but they are all different and nonzero\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "if B = 5, then it's not possible that B+B yields the same digit as B. So C = 0 is not possible. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "This rules out choice A.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "------------------------------------------------------------------\r \n" ); document.write( "\n" ); document.write( "B) 6\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If C = 6, then B has to be 3 or 8. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If B = 3, then B+B=3+3 = 6 but that's not equal to B = 3. So B can't be 3. \n" ); document.write( "If B = 8, then B+B=8+8 = 16. The units digit 6 is not equal to B = 8 (even with a carry). So B can't be 8.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "This rules out choice B\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "-------------------------------------------------------------------\r \n" ); document.write( "\n" ); document.write( "C) 8\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If C = 8, then B has to be 4 or 9\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If B = 4, then B+B doesn't have the units digit of B. So B = 4 is eliminated in this sub-case. \n" ); document.write( "If B = 9, then B+B adds to 9+9 = 18 and we have a carry of 1 from the previous addition. So we really have 19. This makes B+B have a units digit of B. So it fits\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So far, we see that it's possible for C = 8 and B = 9\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "That would make A = 1\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Summary: \n" ); document.write( "A = 1 \n" ); document.write( "B = 9 \n" ); document.write( "C = 8\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Notice how BB+BB = ABC turns into 99+99 = 198 which fits the summary shown above. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Final Answer: C) 8\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Note: we don't need to check D) 9 since we found the answer in choice C.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |