document.write( "Question 126252: I'm stumped and need help with this Algebra problem... \r
\n" ); document.write( "\n" ); document.write( "Find 2 numbers that are 13 times the sum of their digits\r
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\n" ); document.write( "\n" ); document.write( "I tried this but it does not seem to work out... can you help?\r
\n" ); document.write( "\n" ); document.write( " xy = 13(x+y) \n" ); document.write( "

Algebra.Com's Answer #92518 by MathLover1(20850) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "Let $\"M\"$ be the two integer that are respectively 13 times
\n" ); document.write( "the sum of its digits.\r
\n" ); document.write( "\n" ); document.write( "Note that $\"M\"$ $\"MUST\"$ be $\"three-digit\"$ number (the digits are from 0-9, and 9*13=117).\r
\n" ); document.write( "\n" ); document.write( " Let $\"M+=+100a+%2B10b%2Bc\"$ \r
\n" ); document.write( "\n" ); document.write( "Then $\"13%28a+%2B+b%2Bc%29+=+100a+%2B10b%2Bc++\"$ \r
\n" ); document.write( "\n" ); document.write( "We can show that M cannot be a $\"+4-digit\"$ number.
\n" ); document.write( "If M is the number $\"4-digit\"$ number $\"abcd\"$ and all digits are $\"9s\"$ , then\r
\n" ); document.write( "\n" ); document.write( " $\"13+%28a+%2B+b+%2B+c+%2B+d%29+%3C=+13+%2A+9+%2A+4+=+468+%3C+1000\"$ Е.=>Е not possible \r
\n" ); document.write( "\n" ); document.write( "Similarly, larger numbers are not possible.\r
\n" ); document.write( "\n" ); document.write( "We can also show that $\"M+\"$ cannot be a $\"2-digit\"$ number and both digits are $\"9s\"$ . \r
\n" ); document.write( "\n" ); document.write( "If $\"M\"$ is the number $\"+ab\"$ , then\r
\n" ); document.write( "\n" ); document.write( " $\"13+%28a+%2B+b+%29+=+10a%2Bb++\"$ Е.\r
\n" ); document.write( "\n" ); document.write( "=>Е $\"13+a+%2B+13b++=+10a%2Bb++\"$ Е\r
\n" ); document.write( "\n" ); document.write( ".=>Е $\"13+a+%96+10a+%2B+13b+-+b++=+0\"$ Е\r
\n" ); document.write( "\n" ); document.write( ".=>Е $\"3+a+%2B12b++=+0\"$ Е.=>Е \r
\n" ); document.write( "\n" ); document.write( " $\"a+%2B+4b++=+0\"$ \r
\n" ); document.write( "\n" ); document.write( ".=>Е $\"not\"$ possible since $\"a+%3E+0\"$ and $\"b+%3E=+0\"$
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\n" ); document.write( "So the answers can only be $\"3-digit\"$ numbers. Let $\"+M\"$ be $\"abc\"$ , then\r
\n" ); document.write( "\n" ); document.write( " $\"13%28a+%2B+b%2Bc%29+=+100a+%2B10b%2Bc+\"$
\n" ); document.write( "
\n" ); document.write( " $\"13a+%2B+13b+%2B13c+=+100a+%2B+10b+%2B+c\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"0+=87a+-3b+-12c\"$ ЕЕЕЕЕ..divide by $\"3\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"87a%2F3+=3b+%2F3+%2B12c%2F3+\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"29a=b++%2B+4c\"$ \r
\n" ); document.write( "\n" ); document.write( "If $\"a+=+1\"$ , the equation becomes $\"29+=+b+%2B+4c\"$ . We have the following sets of solutions:\r
\n" ); document.write( "\n" ); document.write( " $\"a\"$ | $\"b\"$ | $\"c\"$
\n" ); document.write( " $\"1\"$ | $\"1\"$ | $\"7\"$
\n" ); document.write( " $\"1\"$ | $\"5\"$ | $\"6\"$
\n" ); document.write( " $\"1\"$ | $\"9\"$ | $\"5\"$ \r
\n" ); document.write( "\n" ); document.write( "If $\"a+=+2\"$ , the equation becomes $\"58+=+b+%2B+4c\"$ . This has no solution as the largest value of $\"b+%2B+4c\"$ is $\"45\"$ (when $\"b+=+9\"$ and
\n" ); document.write( " $\"c+=+9\"$ ). \r
\n" ); document.write( "\n" ); document.write( "Similarly, there is no solution for $\"a%3E2\"$ \r
\n" ); document.write( "\n" ); document.write( "Therefore, the possible values of $\"M+\"$ are $\"117\"$ , $\"156\"$ and $\"195\"$ .
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