I've got notified a few days ago from PSE that my account on this website is 1 year old.
So I thought about celebrating it (or commemorating it, because you all made me addicted to this) with a puzzle (or 2).
So here is my alphametic.


This has a unique solution.
After you are done with that, add one more puzzle to the equation and get 2 solutions.



  • every different letter represents a different digit in base 10.
  • the sum is made in base 10
  • Leading zeros are not allowed.
  • equation must be true.
  • an answer without reasoning is not a valid answer.
  • 2
    $\begingroup$ "An answer without reasoning is not a valid answer." - so should this get an no-computers tag? $\endgroup$
    – M Oehm
    Commented Mar 27, 2017 at 7:45
  • $\begingroup$ @MOehm. Right. Thanks. I added it. I always assume that when solving an alphametic computers are prohibited by default. $\endgroup$
    – Marius
    Commented Mar 27, 2017 at 7:47
  • $\begingroup$ @MOehm Couldn't "exhaustive search" be a reasoning? $\endgroup$
    – Brian J
    Commented Mar 27, 2017 at 14:35
  • $\begingroup$ @BrianJ: You could throw all permutations at a computer and get a solution in a second, but that woudn't be much of a puzzle. That's why some puzzles require that you show how you got at the solution under the assumption that you solved it wiothout a computer. $\endgroup$
    – M Oehm
    Commented Mar 27, 2017 at 15:41
  • 1
    $\begingroup$ (Such requirements are okay in small puzzles like this alphametic or logic puzzles along the lines of "The pipe-smoker is neither the person from Caracas nor the one wit the blue socks." But when sudokus require me to document my progress, that's outright tedious. It doesn't even make for good posts, in my opinion.) $\endgroup$
    – M Oehm
    Commented Mar 27, 2017 at 15:44

3 Answers 3


Congratulations on your Yearling!

(Partial answer) For the first puzzle,

E must be a 0 (because E+E=E) and Y a 1.
P+P=10, so P must be a 5.
This leaves us with a simpler alphametic using the numbers 2, 3, 4, 6, 7, 8 and 9:
N is even, and U is 2, 3 or 4. A is at least 4.
Z can't be a 2, because R would be a 4 and O a 4 or 5, which both aren't possible.
Z can't be a 7, because R would be a 5.
Z can't be a 9, because R would be a 9.
Z can't be an 8, because R would be a 7, O a 6, you'd need L=2 and N=4 to fix the last digit (no carrying) and U=3 and A=9 doesn't fit.
Z can't be a 6, because R would be a 3, O a 2, you'd need L=4 and N=8 to fix the last digit (no carrying) and there's no option left for U.
Z can't be a 4, because R would be an 8, O a 9, you'd need L=6 and N=2 to fix the last digit (carrying 1) and U=3 and A=7 doesn't fit.
Therefore, Z is a 3, R is a 6 and O is a 7. For the last digit, we need L=9 and N=8 (carrying 1), and that leaves U=2 and A=4.

The solution is therefore:

  523390 +
  523390 =

  • $\begingroup$ Thanks @boboquack for the edit. I always struggle with formatting inside spoilers, but forgot about good old HTML. $\endgroup$
    – Glorfindel
    Commented Mar 27, 2017 at 8:21
  • $\begingroup$ Good job for the first part. Would you like to take a swing at 3 puzzles sum? $\endgroup$
    – Marius
    Commented Mar 27, 2017 at 8:28

Answer for Part 2,

From the units, E must be either 0 or 5.

Case 1:

E=0. Y can only be 1 or 2, so P must be either 3 (if Y=1), or 6 (if Y=2)

Case 1.1:

E=0, Y=1, P=3. U cannot be 2 because there wouldn't be carryover. U cannot be 7,8,9 because there would be too much carryover (37*3=111). If U=4, then A must be 2 (3 and 4 are already taken) and this is impossible because Z must be at least 5, hence guaranteed to have carryover. If U=5 then A must be either 6 or 7. If A=6 then Z must be 4 to produce 1 carryover, then R must be 3 which is impossible. If A=7 then Z can be 6 (R=0, impossible), 8 (R=6,O=4 but then L and N would be 2 and 9 which is impossible) or 9 (R=9, impossible). So U also cannot be 5. If U=6 then A must be either 8 or 9. If A=8, Z must be 2 which is impossible because R must also be 6. If A=9, Z can be either 4 (R=3, impossible) or 5 (R=6, impossible). This case yields nothing.

Case 1.2:

E=0, Y=2, P=6. U must be 7,8 or 9 to provide enough carryover. U=7 gives A=1 (then Z=3, R=O=9 which is impossible) or A=3 (then Z=8 or 9 which gives R=6 or 9 respectively, both impossible). U=8 gives A=4 (Z=1, R=3, O=5, L=9, N=7 an answer!) or A=5 (Z=3 or 4. Z=3 gives R=0, Z=4 gives R=3 and O=2,3 or 4 which is all taken). U=9 gives A=7 (Z=1, R=3, O=5, L=8, N=4 another answer!) or A=8 (Z=3,4 or 5. Z=3 gives R=0, Z=4 gives R=3 and O=2,3 or 4 which is all taken and Z=5 gives R=6. All impossible. This case as a whole gives 2 answers.


681190 x 3 = 2043570 and 691180 x 3 = 2073540

Case 2:

E=5. If Y=1 then P must also be 5 which is impossible, so Y=2 which yields P=8. U can be 3,4 or 6. If U=3 then A = 0 (Z=4 or 6. Z=4 gives R=3, Z=6 gives R=9 and O=8 or 9. Both impossible. or 1 (Z=6,7 or 9. Z=6 gives R=O=0, Z=7 gives R=3 and O=1,2 or 3 which is all taken. Z=9 gives R=9. All impossible) If U=4 then A=3 which gives Z=6, R=9, O=8 or 9, both taken. If U=6 then A=9 which gives Z=3 or 4. Z=3 gives R=0, O=1 which leaves L,N with 7 or 4 thus not satisfying the equation. Z=4 gives R=3 and O=2,3 or 4 which is all taken. Hence, no solutions from this case.

All cases covered.


Assuming there are no catch, I will go pure logic on this.

Step 1 : E = 0.
Explanation, E + E = E is only true for E = 0.
Result : PUZZL0 + PUZZL0 = Y0ARON0


Step 2 : Y = 1.
Explanation, No number doubled could give an extra 0 and end by a 2 or more.
Result : PUZZL0 + PUZZL0 = 10ARON0


Step 3 : P = 5.
Explanation, only 5 + 5 can give 10.
Result : 5UZZL0 + 5UZZL0 = 10ARON0


-U must be 2,3 or 4 or else the number would end in 11.
-Z must be under 5 or else both Z additions would add up to the same value. So it can be 2,3,4.
-A must therefore be 4,6 or 8
-N is even because there are no carry over from the addition of 0. So it can be 4,6 or 8.
-L must therefore be 2,3,4,7,8 or 9.
-N must therefore be over 4 to make the following Z addition +1. So can be 6,7,8,9.
-R and O can therefore only be (4,6,8)(7,9)
-Z can't be 2 or else one of the 2 Z additions would end in 5.
-Z can't be 4 or else A would be forced to be 6 and R would be 8, so no value left for N.


Step 4 : Z = 3 and therefore R = 6 and O = 7.
Explanation, only possible value left for Z.
Result : 5U33L0 + 5U33L0 = 10A67N0


Only 2,4,8,9 left to use.
Possible combinations for U+U = A are 2-4 and 4-8.
N must be 4 or 8. 4 is impossible because it would block both combinations.


Step 5 : U = 2, A = 4 and N = 8.
Explanation, only possible combination for (U,A,N).
Result : 523390 + 523390 = 1046780

3X puzzle(partial)

E must be 0 or 5.
Y must be 1 or 2.
Z must be 0,1,2,3,7,8,9 or else both Z additions would add to the same value.
P must be 3,4,5,6,7,8 or 9 or else Y would be 0.

  • $\begingroup$ question for the 3x puzzle. Why can't P be 3? I'm not saying it is, but it can be end still end up with Y = 1 if you carry something over from 3x U. $\endgroup$
    – Marius
    Commented Mar 27, 2017 at 9:09
  • $\begingroup$ @Marius oops, you are right. I'll fix that. $\endgroup$ Commented Mar 27, 2017 at 9:37

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