2 ms·

I can fill in some of the next steps... If M=1, then S must equal 8 or 9 since MORE can be at most 1987. Consider O, which is in MORE and MONEY. O cannot be 1

by NeoTar 3y ago

I can fill in some of the next steps... If M=1, then S must equal 8 or 9 since MORE can be at most 1987. Consider O, which is in MORE and MONEY. O cannot be 1 (which is M), and it cannot be 2 or more, since then MONEY - MORE > 10000, so O must be 0 (zero). Considering the hundreds place, E + 0 = E + 0 (zero) = N, so there must be a carry from the previous digit, and N = E + 1 I'm getting a little stuck now, but there must be some more tricks to take it further, presumably since we know N + R > 9 (due to our above carry), etc.

masswerk 3y ago
"S" must be 9, since, if it was 8, "E" had to be 9 + carry, giving "N = 0", which is already occupied. So: 9END 10RE ------- 10NEY Let's have a look at "E + 0 = N", there must be a carry involved, from which follows (considering 0, 1, 9 are already used): N = E + 1, N < 9, E < 8 Considering "N + R = E", which must produce a carry: N + R > 9, E < 6, E > 1, R > 4 So "E" must be in (2, 3, 4, 5), also, substituting "N", there's this interesting notion of (1 +)? E + 1 + R = 10 + E
- tobr 3y ago
  But the question is still… who would work through this in a few seconds?
  - masswerk 3y ago
    I guess, this may become quite natural with practice: eliminating contradictory variants. So it may be well a matter of first glance… (since it's really a matter of tautology and observations like "M = 1" aren't really a matter of thought, as soon as you've encountered this once.) Then you're left with just a few variants to actual think about and maybe try them to see, if they work. Which should become also more effortless with practice. Some of this "second layer" may reveal itself as soon as we've "cracked" the outer shell. So, I guess, you may be able to look at this and say, "well, E must be either 5 or 6 … and, well, it's actually 5", eventually, especially, as well known subproblems become some kind of idiom and we've learned to look for such possible vectors of attack.

by NeoTar 3y ago

masswerk 3y ago

"S" must be 9, since, if it was 8, "E" had to be 9 + carry, giving "N = 0", which is already occupied. So: 9END 10RE ------- 10NEY Let's have a look at "E + 0 = N", there must be a carry involved, from which follows (considering 0, 1, 9 are already used): N = E + 1, N < 9, E < 8 Considering "N + R = E", which must produce a carry: N + R > 9, E < 6, E > 1, R > 4 So "E" must be in (2, 3, 4, 5), also, substituting "N", there's this interesting notion of (1 +)? E + 1 + R = 10 + E

tobr 3y ago
But the question is still… who would work through this in a few seconds?
- masswerk 3y ago
  I guess, this may become quite natural with practice: eliminating contradictory variants. So it may be well a matter of first glance… (since it's really a matter of tautology and observations like "M = 1" aren't really a matter of thought, as soon as you've encountered this once.) Then you're left with just a few variants to actual think about and maybe try them to see, if they work. Which should become also more effortless with practice. Some of this "second layer" may reveal itself as soon as we've "cracked" the outer shell. So, I guess, you may be able to look at this and say, "well, E must be either 5 or 6 … and, well, it's actually 5", eventually, especially, as well known subproblems become some kind of idiom and we've learned to look for such possible vectors of attack.