document.write( "Question 465888: What is the smallest number that increases its value by 12 when the number's order of the digits is reversed? \n" ); document.write( "
Algebra.Com's Answer #319302 by richard1234(7193)\"\" \"About 
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I think I've already solved this problem; there is no such positive integer. This is because the digit sum remains the same, so the remainder upon dividing by 9 will be the same.\r
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\n" ); document.write( "\n" ); document.write( "To see this, take some random numbers like 583, 79, and 1056. Reversing their digits we get 385, 97, and 6501. If we take their differences we get\r
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\n" ); document.write( "\n" ); document.write( "583 - 385 = 198 = 22*9
\n" ); document.write( "79 - 97 = -18 = -2*9
\n" ); document.write( "1056 - 6501 = -5445 = -605*9\r
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\n" ); document.write( "\n" ); document.write( "In all these cases we are essentially adding or subtracting multiples of 9, and since 12 is not a multiple of 9 we cannot make a number increase by 12.\r
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\n" ); document.write( "\n" ); document.write( "However this is entirely possible in other bases. If we have (12 in base 13 = 15) and we reverse the digits we get , and\r
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\n" ); document.write( "\n" ); document.write( "See if you can find the smallest base-10 number that, when written in a different base, the property holds (here the base-10 number was 15).
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