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In Round 206, the 'Lenny Conundrum' authors, rather unoriginally, wrote:Assume each letter in the following formula represents a unique digit: (for example, ABCD could be the number 1234)
NOMAD x 4 = DAMON
What is the number represented by the digits DAMON?
You are multiplying a five-digit number and getting a five-digit number. This means that there is no "carry" from the ten-thousands' place. Since 4 ?- 3 = 12, and very larger multiple is two digits as well, this limits the leading N in NOMAD to being 1 or 2, to prevent the (unwanted) "carry".
Since D ?- 4 = N (or N + 10, if there is a "carry" to the tens' place), you have:
. . . . .4 ?- 1 = 4
. . . . .4 ?- 2 = 8
. . . . .4 ?- 3 = 12
. . . . .4 ?- 4 = 16
. . . . .4 ?- 5 = 20
. . . . .4 ?- 6 = 24
. . . . .4 ?- 7 = 28
. . . . .4 ?- 8 = 32
. . . . .4 ?- 9 = 36
(Obviously D cannot be zero, since then N would also be zero, and leading zeroes are not allowed in whole numbers.)
So D ?- 4 = N ends in any of 2, 4, 6, and 8. Since we already know that N must be 1 or 2, we must have N = 2, and thus D = 3 or D = 8.
. . . . .2 O M A D ?- 4 = D A M O 2
Since N = 2 and since 4 ?- 2 = 8, then D must be 8:
. . . . .2 O M A 8 ?- 4 = 8 A M O 2
Since 4 ?- 8 = 32, there will be a "carry" of 3 into the tens' place. We can see that there cannot be a "carry" from the O, because then D would be more than 8. So O must be 1 or 2. Suppose O = 1. Then:
. . . . .2 1 M A 8 ?- 4 = 8 A M 1 2
From the ones' place, we have 4 ?- 8 = 32, so there is a "carry" of 3 into the tens' place. This means that 4 ?- A + 3 = ??, where ?? has a ones' digit of 1 (from O).
Suppose ?? = 11. Then 4 ?- A = 8, so A = 2. But 2 is already taken.
Suppose ?? = 21. Then 4 ?- A = 18, so A = 18/4 = 9/2. But this is not possible.
Suppose ?? = 31. Then 4 ?- A = 28, so A = 7:
. . . . .2 1 M 7 8 ?- 4 = 8 7 M 1 2
So far, we have:
. . . . .(21000 + M + 78) ?- 4 = 84000 + 4M + 312 = 84312 + 4M
Then we need a "carry" of 3 from 4M. We know that, with the "carry" from the ten's place, the result of the hundreds' place multiplication will be 4 ?- M + 3 = 30 + #. Then 4 ?- M is at least 27, so M is at least 7.
Since 7 and 8 are already taken, then M must be 9 (assuming our "suppose O = 1" was correct). Then:
. . . . .2 1 9 7 8 ?- 4
Since 21978 ?- 4 = 87912, then these are the correct values.
Eliz.