Neat
Posted: November 12th, 2019, 11:20 am
by cinelli
Here is a quick alphametic.
ANTE
ETNA -
----
NEAT
Cinelli
Re: Neat
Posted: November 12th, 2019, 11:46 pm
by UncleEbenezer
Reversing the digits and subtracting reminds me of your recent puzzle.
And lo, revisiting that thread, I see you spelled out, 7641 - 1467 = 6174
a = 7, n = 6, t = 4, e = 1
Was that intentional?
Re: Neat
Posted: November 13th, 2019, 11:32 pm
by Gengulphus
cinelli wrote:Here is a quick alphametic.
ANTE
ETNA -
----
NEAT
Spoiler...
It's a bit easier to think about this if we turn it into the addition:
bcd
NEAT
ETNA +
----
ANTE
in which c, d and e are the carries between the columns, each equal to 0 or 1. The individual column sums are then:
b+N+E = A
c+E+T = N+10b
d+A+N = T+10c
T+A = E+10d
Adding the first and last of those together gives b+N+E+T+A = A+E+10d, or b+N+T = 10d. Since b >= 0, N >= 0, T >= 0 and N and T differ, that implies that 10d > 0, so d=1 and b+N+T = 10, or N = 10-T-b.
Substituting N = 10-T-b in the second of them gives c+E+T = 10-T+9b, or E = 10-2T+9b-c.
Substituting E = 10-2T+9b-c and d=1 in the last of them gives T+A = 10-2T+9b-c+10, or A = 20-3T+9b-c.
Substituting d=1, A = 20-3T+9b-c and N = 10-T-b in the third of them gives 31-4T+8b-c = T+10c, or 31+8b-11c = 5T. Since b and c must each be 0 or 1, that gives 5T = 20, 28, 31 or 39, among which only 20 is a multiple of 5. So 5T = 20, b=0 and c=1, and it follows that T = 4, N = 10-T-b = 6, E = 10-2T+9b-c = 1 and A = 20-3T+9b-c = 7:
7641
1467 -
----
6174
Gengulphus
Re: Neat
Posted: November 14th, 2019, 9:12 pm
by cinelli
Reply to spoilers:
Well solved both.
UncleEbenezer wrote:And lo, revisiting that thread, I see you spelled out, 7641 - 1467 = 6174
Was that intentional?
It was a coincidence, actually. But 6174 is my favourite number. 
Cinelli