# How to find the value that makes this addition equation work without using a computer?

The following image represents the addition of 2 five-digit numbers, where 10 different letters represent 10 different numbers (0~9). What does PUZZLE equal? (There are two possibilities and a complete answer is required) source I can only solve this by using a computer. The answer is 34752+75234=109986 and 53927+92753=146680. Does anyone know how to solve it with only pen and paper, no computers?

• ZZ is the key with 4 different numbers RANT.
– Oray
Feb 12 at 9:26
• @Oray I don't see what you mean. For Z=0 alone I find 40 possibilities for R,A,N,T. Feb 12 at 10:32
• The existence of multiple solutions excludes the possibility of logical deduction. Feb 12 at 14:00
• One thing you can use is that the ten digits are split between the terms and the sum and further that the two terms have both the exact same digits. By considering digit sums this leads to the identity:2 x (T+R+A+I+N) = (P+U+Z+L+E) + Z mod 9, hence 3 x (T+R+A+I+N) = Z mod 9 so Z can only be one of 0,3,6,9 Feb 12 at 15:21
• @daw Why do you say that isn't a solution? The sum is correct and each letter stands for a distinct digit consistently. (Admittedly, it is written as INTRA+TRAIN, which is backward compared to the image, but that doesn't affect its validity.)
– fljx
Feb 13 at 16:06

UPDATE Solution is now technically complete.

Here is a sketch:

We will use the

Fact: the digit sum of a sum is the sum of the digit sums minus 9 times the number of carries c.

Let us write u = T+R+A+I+N and x = P+U+Z+L+E. Then u + x = 45. Applying the Fact to the given sum yields 2u - 9c = x + Z or 3u = 45 + Z + 9c. In particular, Z is a multiple of 3.

This is case-bashable, but rather ugly: Each of the four cases Z=0,3,6,9 has, in principle, fivefour subcases c=1,...,4 UPDATE there cannot be five carries because whichever letter happens to be 9 rules out at least one carry /UPDATE but many of those can be ruled out off the bat. Each subcase has u fully determined plus some constraints on R+N and A+T.

Case Z=0:

The columns with a Z in them must both give a carry, therefore T,I,R,N cannot be 9 nor can A because it must sum to 9 or 10 with T and T can be neither 0 nor 1. But this leaves almost no room for 9: 9 = R+N != N+A = E rules out E while P and U are, of course, not possible either. This only leaves L and requires L to to receive and not give a carry. It follows c=4, u=27 I=8, R=0, contradiction.

Case Z=3:

a) both Z columns trigger a carry: then u=25 or u=28. Also, R+N+A+T=24 or 25 ruling out u=25 as I cannot be 0 or 1. But u=28 leaves only I=4 which together with Z=3 blocks two of the three ways to make 12 (9+3,8+4,7+5) leaving not enough for R+N=A+T=12.

b) there is a Z without carry. First, we observe that there can be only one and it has to receive a carry and it must be 0+2. As only R can be 0 this implies c=3 => u=25,R+M+A+T=15, i=10. Contradictin.

Case Z=6:

Then there must be one carrying and one non carrying Z.

a) The carrying Z also receives a carry: Then A,T= 7,8, R,N=2,3 But then for I+R to yield a carry I has to be 9 and PU=I+T=16 or 17 contradiction.

b) The carrying Z receives no carry: Then it is 9+7. b1) A,T=9,7. Then N>=2, therefore N+A must carry => c=3 and u=26, R+N+A+T=21, I=5, R,N=2,3, T=9, U=4, solution: 92753+53927=146680 b2) R,N=9,7. Then I must be 8 or 5 therefore u=21, A,T=2,3 but the last entails U=1 or 2, contradiction.

Case Z=9:

Then neither column with a Z in them can give a carry (the right one could in principle receive one.) This leaves subcases c=1,2,3 but c=3 would have R+N+A+T=Z+Z-1=17 and u=27 hence I=10 which is not allowed. c=2 has I=6 or I=7 depending on whether the second carry happens in the right most column or the one next to it. Note that in either case we can pinpoint the location of 8 by elimination L=8 if I=6, E=8 if I=7. But the first of the two implies R=1 contradicting P=1. The second implies A+T = N+A = 8, also a contradiction.

This leaves c=1. Then I=3 => T=7 => U=0,A=2,R,N=4,5 => E=6,L=8, solution 75234+34752=109986

This case-bashing is still rather tedious, though, so there is room for improvement.

• Also P=1. so none of the other letters can equal 1, reducing the subcases. Feb 13 at 8:30
• TRA*101 + IN*1001 = 1UZZLE => TRA*2 = 1UZZLE (mod 11) => 2 * (T+A-R) = E+U - (1+L) (mod 11), TRA*3= 1UZZLE (mod 7) => 3 * (2T+A+3R) = E+3L+Z+4U+5 (mod 7), N+A = 10n + E, and TRA*10=1UZZLE (mod 13) => 10*(A-3R+9T) = E-3L+8Z+3U+ 4 (mod 13) Feb 13 at 10:46
• TY! Why R+N=9 and L = I+R != R+N != N+A = E rules out L and E when Z=0? Is it impossible e.g. I+R = 8 and get a carry from N+A? (sry if it sounds dumb) Feb 13 at 17:52
• @SeniorTame I guess you're right, I'll have to find a better argument there. Feb 13 at 18:19