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Alphametic


Go to solution Solved by EventHorizon,

Question

OK these subtractions do not work using letters.
But make a number substitution, the same for both,
and everything's fine.

N I N E             N I N E
- T E N             - O N E
=======             =======
  T W O               A L L

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  • Solution
Spoiler

L=0 (since E-E=L)

N=1 (since the largest two three-digit numbers can sum to is 1998)

A+O = 10 + I (A+O need to have a carry of 1 to make the 4 digit number)

O+1=E

W+E=11 (since N is 1, W+E would be too much so we need to borrow from the next)

2*T = I + 9 (since we borrowed above and the 1000's place has a 1), also I is odd.

 

O+1=E and W+E=11 => W+O=10

A+O = (W+O) + I => A=W+I

 

Since I is odd and the smallest W could be is 2, I is at most 7, but could be 5 or 3, too.

 

Assume I = 7.  2T=9+7 = 16, so T=8.  A=W+I => A=W+7, so A=9 and W=2.  10=W+O => 10=2+O => O=8, same as T.

 

Assume I=5.  T=(9+5)/2=7

---> Assume A=9.  9=W+5 => W=4.  10=4+O => O=6.  E=6+1=7, same as T

---> Assume A=8.  8=W+5 => W=3.  10=3+O => O=7, same as T

---> A can't be 7 (T is), and W needs to be at least 2 (0 and 1 are taken), so done here.

 

Assume I=3.  T=(9+3)/2=6

---> A can't be 9, it would make W = 6, same as T.  A can't be 6, it would make W=I.  A can be 8,7,or 5.

---> Assume A=8.  8=W+3 => W=5.  10=5+O => O=5, same as W

---> Assume A=7.  7=W+3 => W=4.  10=4+O => O=6, same as T

---> Assume A=5.  5=W+3 => W=2.  10=2+O => O=8.  E=8+1=9.  Looks like this is the only solution.

So L=0, N=1, W=2, I=3, A=5, T=6, O=8, E=9.

NINE-TEN=TWO => 1319-691=628

NINE-ONE=ALL => 1319-819=500

 

 

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