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# Solution

Well, this is how I worked it out. It took me hours. I am sure you have a shorter method and must have solved it within minutes.

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
because the sum of two digits is less than 20 (for instance, even if both the digits are 9, their sum will be 18. If there is a carry over
of 1 from the addition of the previous two digits, the sum will at the most be 19. The left-most digit is T and it is 1). So, **T = 1**.
It is already given that **A = 6**.

Now consider the addition of the right-most digits. The addition of E and N may or may not have a carry. Suppose there is no carry, then E + N = D. In that case, we proceed to the next two digits R and I. The sum of R and I may or may not have a carry; so, R + I = I or R + I = 10 + I. If R + I = I, then R = 0. This is not possible because R, C, and T are not zero. So, R + I = 10 + I. But, this implies that R = 10, which again is not possible because R is a single-digit number. Therefore, one thing is certain: the sum of E and N has a carry. Therefore, E + N = 10 + D.

This implies that 1 + R + I = I (if this sum doesn't have a carry) or 1 + R + I = 10 + I (if this sum 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; **R = 9**. So, the
addition of R and I has a carry.

Since there is a carry from R + I, we have 1 + A + O = P (the addition of A and O doesn't have a carry) or 1 + A + O = 10 + P (the addition of A and O has a carry). Now, A = 6. Therefore, 1 + 6 + O = P or 1 + 6 + O = 10 + P. That is, 7 + O = P or 7 + O = 10 + P.

Let's first consider the second case which has a carry: 7 + O = 10 + P. The carry from the addition of two digits is always 1, so this 1 will be carried over when adding the next two digits R and C. So, 1 + R + C = E (this addition doesn't have a carry) or 1 + R + C = 10 + E (this addition has a carry). We know for certain there is a carry, otherwise 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 because the digits are distinct. This difficulty has arisen because we considered 7 + O = 10 + P to be true. Thus, 7 + O = 10 + P is false and 7 + O = P is true.

So we have established that 7 + O = P. This means the addition of A and O does not have a carry. 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) which is P = 8 and O = 1. But 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 P cannot be 9. Therefore, (9,2) is also ruled out. That leaves us with only (7,0).
Therefore **P = 7 and O = 0.**

Phew!

So far, we have found that **A = 6, R = 9, O = 0, T = 1 and P = 7**. That leaves us with 2, 3, 4, 5, and 8; the rest of the alphabets can be from
these five digits.

We have seen that addition of A and O doesn't have a carry, therefore 1 + R + C = 10 + E is not true. R + C = 10 + E is true. Therefore, 9 + C = 10 + E.
Therefore, C - E = 1. The (C,E) pairs from the remaining digits which satisfy this are (5,4), (4,3) and (3,2). Therefore C can be 5, 4, or 3 and E can be 4, 3, or 2.
We have seen that 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**. The only remaining digit 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.

**Some useful links for your career:**

- Union Public Service Commission - www.upsc.gov.in
- IIT-Kharagpur - www.iitkgp.ac.in
- Indian Statistical Institute - www.isical.ac.in
- Indian Institute of Technology Madras - www.iitm.ac.in
- Indian Institute of Management, Ahmedabad - www.iimahd.ernet.in
- Indian Institute of Mass Commission - www.iimc.nic.in
- IIT Bombay - www.iitb.ac.in
- Indian School of Mines, Dhanbad - www.ismdhanbad.ac.in
- Birla Institute of Technology, Ranchi - www.bitmesra.ac.in
- Central Institute of Fisheries Nautical and Engineering Training - www.cifnet.nic.in
- Indian Institute of Information Technology, Allahabad (Deemed University) - www.iiita.ac.in
- Central Marine Fisheries Research Institute, Kochi - www.cmfri.com
- Tata Institute of Social Sciences, Mumbai - www.tiss.edu