Presumably you insist on at least one "+" on the LHS --- otherwise a trivial solution is 9876543210 ?! WFL On 1/21/12, David Wilson <davidwwilson@comcast.net> wrote:

Consider equations like the following:

0+4+8+27+96=135

with an addition expression on the left side and a sum on the right side, in which each digit occurs exactly once.

If you don't like 0 as a term, you can usually find another expression with same sum value:

2+6+9+48+70=135

Suppose, however, that the 0 term cannot be eliminated.

What is the sum?

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I haven't seen a real answer yet, so here's a spoiler... The 0 can always be eliminated if there is a multi-digit number in the sum (for example, 0 + xyz can be replaced by z + xy0). So to prevent such a transformation, it must be a sum of all single-digit numbers. To make this work, the sum plus the sum of its digits must be 45. Only 36 fits this requirement. But that is not sufficient--it must also be shown that 36 cannot be the sum involving a two-digit number, otherwise there would be no solution. But this is not a problem since the minimum sum with a two-digit number (and without a 3 or 6) is 45. Thus the solution is 0+1+2+4+5+7+8+9=36. Thanks for the "mild" (i.e. logic instead of search) puzzle! Nick On 1/20/2012 8:36 PM, David Wilson wrote:

Consider equations like the following:

0+4+8+27+96=135

with an addition expression on the left side and a sum on the right side, in which each digit occurs exactly once.

If you don't like 0 as a term, you can usually find another expression with same sum value:

2+6+9+48+70=135

Suppose, however, that the 0 term cannot be eliminated.

What is the sum?

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