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.
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.
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.
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.
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.
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 9so Z can only be one of 0,3,6,9 $\endgroup$