The equation MATH - COU = NTS can be made true if each of the letters M,A,T,H,C,O,U,N and S is replaced by a different digit and the letter groups are read as 4 digit and 3 digit integers. Both Ts have the same value. If C = 3, what is the value of the letter O?
So far I only have figure that you can use numbers 1-9. But where to go from there...?
If the numbers 1 - 9 are the only choices:
You will find it easier if you convert the problem to an addition problem as in COU + NTS = MATH
.COU
.NTS
MATH
If C = 3
.3OU
.NTS
MATH
The first, and easiest conclusion you can immediately make is that M must be 1 as the sum of 3 and N can be no more than 14, i.e., 3 + 9 + 1 carryover from O + T. So we now have
.3OU
.NTS
1ATH
N can only be 8 or 9.
If N = 6, 3 + 6 is less than 10.
If N = 6 with a 1 carryover from O + T, 3 + 6 += 1 = 10, but 0 is not a choice.
If N = 7, 3 + 7 = 10, and , again, 0 is not a choice.
If n = 7 with a 1 carryover from O + T, 3 + 7 + 1 = 11, but M already = 1.
If N = 8, 3 + 8 = 11, but, again, m already = 1.
If N = 8 with a 1 carryover, 3 + 8 + 1 = 12, making A = 2.
.3OU
.8TS
12TH
Here , we are faced with a dilemma.
O + T = T requires that 1) O = 0, but 0 is not a choice or 2) O = 9 and T = 4, 5, 6 or 7 with a 1 carryover from U + S.
Why? 9 + any of these numbers are the only remaining numbers resulting in the 2 T's being the same number. 9 + 4 + 1 = 14; 9 + 5 + 1 = 15; 9 + 6 + 1 = 15; 9 + 7 + 1 = 17.
Can you take it to a final conclusion?
If you find no solution, then perhaps the number 0 must be a possible choice somewhere.
What other values of O + T can result in O + T = T?
