January 2013

# Solution

Well, this is how I worked it out. It took me hours. I am sure you have a shorter method and must have worked it out in minutes. I presume you are here only to cross-check your answer.

First of all, since the sum of two four-digit numbers is a five-digit number, one thing is certain - "T" is a carry over and it can be 1 and nothing else (the carry cannot be greater than or equal to 2). So, T = 1.

Now consider the addition of the left-most digits, either the addition of "E" and "N" will have a carry or not. Suppose there is no carry, then E + N = D, which means R + I = I or R + I = 10 + I (that is the addition has a carry over). If R + I = I, then R = 0 which is not possible. In case R + I = 10 + I, then R = 10 which again is not possible because "R" is a single-digit number. Therefore, one thing is sure - the addition of "E" and "N" has a carry. Therefore, E + N = 10 + D.

This means that 1 + R + I = I (the addition does not have a carry) or 1 + R + I = 10 + I (the addition has a carry). In the first case, 1 + R + I = I. Then, 1 + R = 0. Therefore, R = -1 which is not possible. In the second case, 1 + R + I = 10 + I. Therefore, 1 + R = 10 and R = 9.

Since there is a carry from R + I, we have 1 + A + O = P or 1 + A + O = 10 + P (this addition has a carry). Now A = 6 (given). Therefore, 7 + O = P or 7 + O = 10 + P. Now if 1 + A + O = 10 + P (which means there is a carry), then 1 + R + C = E or 1 + R + C = 10 + E. We know for sure there is a carry, else we will not have "T". So 1 + R + C = E is false and 1 + R + C = 10 + E is true. But R = 9. Therefore, 10 + C = 10 + E which means C = E. This is not possible. Therefore, 1 + A + O = 10 + P is false and 1 + A + O = P is true. That is 7 + O = P (Because A = 6).

So we have established that 7 + O = P. Therefore, P - O = 7. The (P,O) pairs which satisfy this are: (7,0), (8,1) and (9,2) - Since we are dealing with single digits here, pairs like (10, 3), (11,4), etc are not the right choices.. Consider the pair (8,1). We have already established that T = 1. There cannot be two alphabets with the same value. So (8,1) can be ruled out. Likewise R = 9, so (9,2) can also be ruled out. That leaves us with only (7,0). Therefore P = 7 and O = 0.

Phew!

We have found that A = 6, R = 9, O = 0, T = 1 and P = 7. So we are left with the numbers (2, 3, 4, 5, 8). We have seen that if 1 + R + C = 10 + E, then C = E which is not possible. So, R + C = 10 + E. Therefore, 9 + C = 10 + E. Therefore, C - E = 1. The (C,E) pairs from the remaining numbers which satisfy this are (5,4), (4,3) and (3,2). Again, E + N = 10 + D. Let us replace E by 4, 3 and 2 and see what happens. When E = 4, we get N - D = 6. Only (8,2) satisfies this. When E = 3, N - D = 7. There is no pair which satisfies this. Finally, when E = 2, N - D = 8. Again there is no pair which satisfies this. Therefore, (8,2) for the (N,D) pair when E = 4 is true. Thus, E = 4, N = 8 and D = 2.

Phew!

Finally, R + C = 10 + E. Therefore, 9 + C = 10 + 4. Therefore C = 5 and the only remaining number is 3 and the only remaining alphabet is I. Therefore, I = 3.

Oh my god, I have pulled all the hairs from my head. What we finally have is: 9694 + 5038 = 14732.

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