document.write( "Question 1123434:
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Algebra.Com's Answer #739759 by solver91311(24713) $\"\"$ $\"About$

You can put this solution on YOUR website!
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document.write( "      WE\r\n" );
document.write( "     NOW\r\n" );
document.write( " +  KNOW\r\n" );
document.write( "   ELLIE\r\n" );
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\n" ); document.write( "\n" ); document.write( "Since adding nothing to K results in a different number, there must be a carry from the addition of N + N. Since K plus a carry is a two-digit number and the largest possible result of adding two different single digits one of which is at most 2 is 11, E must be 1.\r
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\n" ); document.write( "\n" ); document.write( "In order for the carry from N + N + a carry to be as much as 2, N must be 9, which would mean that K cannot be 9, it must be 8, so the result would be that L = 0. If the carry from N + N + a carry is one, then K must be 9 so that a two-digit result is obtained from K + a carry, and again L = 0.\r
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\n" ); document.write( "\n" ); document.write( "Since 1 + W + W = 1 plus a possible carry, W must be either 0 or 5. But L is already 0, so W = 5.\r
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\n" ); document.write( "\n" ); document.write( "So far we have:\r
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document.write( "      51\r\n" );
document.write( "     NO5\r\n" );
document.write( " +  KNO5\r\n" );
document.write( "   100I1\r\n" );
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\r
\n" ); document.write( "\n" ); document.write( "Let's see what happens if we assume K = 8. In that case, the carry from N + N + a carry must be 2, and so N + N + a carry must be equal to 20, which means N = 9, and the carry from 5 + O + O + a carry must be 2. That would give us:\r
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document.write( "      51\r\n" );
document.write( "     9O5\r\n" );
document.write( " +  89O5\r\n" );
document.write( "   100I1\r\n" );
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\n" ); document.write( "\n" ); document.write( "But in order for 5 + O + O to be greater than 20, O would have to be either 8 or 9, and both of those are used. Therefore, K = 9. And we have:\r
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document.write( "      51\r\n" );
document.write( "     NO5\r\n" );
document.write( " +  9NO5\r\n" );
document.write( "   100I1\r\n" );
document.write( "

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\n" ); document.write( "\n" ); document.write( "From this we see clearly that N + N + a carry must equal 10. But we also know that the carry must be 2 because N + N is perforce an even number. So if the carry is 2, N + N = 8, which is to say N = 4.\r
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document.write( "      51\r\n" );
document.write( "     4O5\r\n" );
document.write( " +  94O5\r\n" );
document.write( "   100I1\r\n" );
document.write( "

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\n" ); document.write( "\n" ); document.write( "And we are left with finding O and I. Clearly, the carry from 1 + 5 + 5 is 1, and we know that 5 + O + O + 1 is larger than 20. If O is 6 or less, the sum is less than 20. If O is 7 then the sum is 20, but that would make the value of I = 0, but 0 is already assigned to L. So O must be 8 or 9, but 9 is already used. O = 8 and since 5 + 1 + 8 + 8 = 22, I = 2.\r
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document.write( "      51\r\n" );
document.write( "     485\r\n" );
document.write( " +  9485\r\n" );
document.write( "   10021\r\n" );
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\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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