This is my (constructive) way of protesting the new sandbox for riddles.
It's a simple alphametic.
You know the rules.
Each letter is a unique digit.
The leading digits cannot be 0 (zero).
If you offer a solution, please add your reasoning.


Since A & B and X & D can be switched and still keep a valid result, this one has 4 solutions.
I will settle for one.

  • 6
    $\begingroup$ Disagreeing with your protest about the sandbox but totally agreeing with your way of protesting against it. +1. $\endgroup$
    – IAmInPLS
    Commented Aug 24, 2016 at 7:55
  • 1
    $\begingroup$ @IAmInPLS. my opinion is this is how all protests should be. everywhere. A constructive way against what you don't agree with. Not a destructive one. Or at least a passive-aggressive approach instead of an aggressive-aggressive one. $\endgroup$
    – Marius
    Commented Aug 24, 2016 at 8:01
  • $\begingroup$ I am all for the sandbox, its a great idea in my mind. Still, i absolutely love this post! nice job. +1 $\endgroup$
    – Aric
    Commented Aug 24, 2016 at 10:11
  • 1
    $\begingroup$ This is how protests should be carried out! :) $\endgroup$ Commented Aug 24, 2016 at 10:55
  • $\begingroup$ @Marius You're lucky: back in the day, Jon Ericson threatened to suspend anyone who posted meta-commentary about site policy in puzzles ;-) $\endgroup$ Commented Aug 31, 2016 at 13:02

3 Answers 3


S must be 9 and N=1 with O=0. Then D+X must carry (because N+O=U can't be 1), so U=2.
Which leaves 3,4,5,6,7,8
D+X=11, and A+B=F which must carry and F>2, so A/B are 58,68,78,67, which gives:
D,X=4,7, A,B,F=5,8,3 (9514+807=10321)
68 means both F and D/X are 4, 78 uses 5 twice and 67 uses 3 twice.


\begin{align}9514\\+\quad807\\\hline10321\end{align} As usual, $\text{N}$ must be $1$, because $\text{SAND} \le 9999$ and $\text{BOX} \le 999$, so $\text{NOFUN}$ cannot be more than $19999$.  In fact, it can't be more than $10998$, so $\text{O}$ (letter oh) is $0$ (zero).  From the ones' column, $\text{D+X=11}$, so we carry $1$ into the tens' column and get $\text{U=2}$.

$\text{D}$ and $\text{X}$ can be $3+8$, $4+7$, or $5+6$.  From the hundreds' column, $\text{A+B=10+F}$ (because we need to carry into the thousands' column).  Viable options are $5+8=13$, $6+7=13$, $6+8=14$, and $7+8=15$.  Of those, all but the first eliminate all possibilities for $\text{D}$ and $\text{X}$, so we must have $\text{A}$ and $\text{B}$ $=$ $5$ and $8$.  That eliminates the first and third options for $\text{D}$ and $\text{X}$, leaving us with $4$ and $7$.

$\text{A}$ and $\text{B}$ $=$ $5$ and $8$
$\text{D}$ and $\text{X}$ $=$ $4$ and $7$

  • $\begingroup$ You know the rules for alphametics. A simple solution with no reasoning is not a solution at all. There are a lot of alphametic solvers online. this does not show that you put some effort in it. $\endgroup$
    – Marius
    Commented Aug 24, 2016 at 7:12
  • $\begingroup$ Fast answer and very clear explanation, nice work! $\endgroup$
    – elias
    Commented Aug 24, 2016 at 7:39
  • $\begingroup$ this is a nice explanation, but a full answer with explanation was added before this one. just +1. $\endgroup$
    – Marius
    Commented Aug 24, 2016 at 7:59

On the first look you get that

S = 9, otherwise you wouldn't get a 5 digit sum
That means that N = 1 and O = 0

So now we get this:


Let's dig deeper:

D + X can't be 1 (zero is already taken), so D + X = 11
Now we know that U = 2 (because of D + X = 11)
We also see that A + B > 10, that means that F = A + B - 10

The letters and numbers left are:

A, B, F, D, X and 3, 4, 5, 6, 7, 8.
D and X are either 3 and 8, or 4 and 7, or 5 and 6.
A + B - 10 must give a number listed above. Trial and error skills:

We got that A and B are 5 and 8 (using another possibility would take up all valid pairs of numbers for D and X).
A + B = 13. F = 3

The only remaining pair for D and X is 4 and 7.

So, altogether we get:

4 solutions. I'll write one here:



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