document.write( "Question 392999: when a three digit number is divided by the sum of the digits of the number, the quotient is 26. What is the least number for which this is true? \n" ); document.write( "
Algebra.Com's Answer #278965 by richard1234(7193)\"\" \"About 
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We have\r
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\n" ); document.write( "\n" ); document.write( "\"%28100a+%2B+10b+%2B+c%29%2F%28a+%2B+b+%2B+c%29+=+26\" --> \"100a+%2B+10b+%2B+c+=+26a+%2B+26b+%2B+26c\" (a, b, c are digits). Subtracting \"a%2Bb%2Bc\" from both sides,\r
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\n" ); document.write( "\n" ); document.write( "\"99a+%2B+9b+=+25a+%2B+25b+%2B+25c\"\r
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\n" ); document.write( "\n" ); document.write( "This implies 9 divides a+b+c and 25 divides 11a+b. Therefore, the only possible values of a+b+c are 9, 18, 27 (since 36 is too high).\r
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\n" ); document.write( "\n" ); document.write( "Suppose a+b+c = 9. Then, \"a%2Bb+=+9-c\", and \"11a+%2B+b+=+9+-+c+%2B+10a\" (by substituting) which is divisible by 25. Since 9 - c + 10a is congruent to 9 - c (mod 10) = 5 or 0, the only possibility for c is 4 (since 9 is impossible). This leaves \"a%2Bb+=+5\" (note that 11a + b is divisible by 25). The easiest way is simply guess and check to find any solutions; we see that a = 2, b = 3 works since 11(2) + 3 is divisible by 25.\r
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\n" ); document.write( "\n" ); document.write( "Therefore the smallest number is 234.
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