The French mathematician Lagrange proved that every natural number can be written as the sum of four squares.

This is true in the view of cryptarithms, too.


Each letter represents a unique digit.

  • $\begingroup$ This should be shifted to Math.SE ... $\endgroup$
    – pranav
    Feb 2 '15 at 16:08
  • 3
    $\begingroup$ @pranav Why? What would be the purpose of the [verbal-arithmetic] tag, then? (Note that this isn't actually asking for a proof of anything; as I understand it, SQUARE and NUMBER are just six-digit strings, like any other cryptarithm.) $\endgroup$
    – wchargin
    Feb 2 '15 at 20:45
  • 2
    $\begingroup$ I think @pranav was joking. :-) $\endgroup$
    – P.-S. Park
    Feb 3 '15 at 12:45
  • $\begingroup$ @P.-S.Park , :) $\endgroup$
    – pranav
    Feb 3 '15 at 16:16

From the last two positions, we see that $4\cdot(10R+E)\equiv 10E+R \bmod{100}$. This is equivalent to $39R\equiv6E\bmod{100}$, and further to $13R\equiv2E\bmod{100}$. Hence $R$ has to be even. Checking $R=0,2,4,6,8$, we see that only $R=8$ works.

We conclude that $R=8$ and that $E=2$, and that the carry over from the second column to the third column is $3$.

Since $4\cdot10^5\,S= 4\cdot S00000 < 4*\mbox{SQUARE} = \mbox{NUMBER} <10^6$, we conclude $4S<10$. The case $S=0$ would have zero as leading digit, and the case $S=2$ would collide with $S=E=2$.

We derive $S=1$.

We plug the detected values into the given equation and derive $4\cdot(1000+100Q+10U+A)+3=1000N+100U+10M+B$,
which simplifies to
$4003 -10M+ 4A-B = 1000N -400Q +60U.~~~ (*)$

The left hand side of equation ($*$) is at least $4003-90+0-9=3904$ and at most $4003-0+36-0=4039$. Furthermore, it is a multiple of $20$. Hence it lies between $3920$ and $4020$, which yields $196\le 50N-20Q+3U\le 201~~~ (**)$.

Now we distinguish several cases on $U$ to derive the following from $(**)$:

(1) If $U=0$, then $5N-2Q=20$. Then $N=6$ and $Q=5$.
(2) If $U=3$, then $5N-2Q=19$. Then $(N,Q)=(5,2)$ or $=(7,8)$; collisions.
(3) If $U\in\{4,5\}$, then $181<50N-20Q<189$. No solution.
(4) If $U\in\{6,7\}$, then $5N-2Q=18$. Then $(N,Q)=(4,1)$ or $=(6,6)$; collisions.
(5) If $U=9$, then $5N-2Q=17$. Then $(N,Q)=(5,4)$ which is fine, or $(N,Q)=(7,9)$ which has the collision $Q=U=9$.

All in all, this leaves us with two possible cases:

Case X: $(N,Q,U)=(6,5,0)$
Case Y: $(N,Q,U)=(5,4,9)$

In case X, equation $(*)$ boils down to $10M- 4A+B = 3$.

  • If $M\ge4$, then $10M-4A+B\ge40-4\cdot9+0=4>3$; contradiction.
  • If $M=3$, then $4A-B=27$ implies $(A,B)=(7,1)$ or $=(8,5)$ or $=(9,9)$; all three cases are collisions (with $B=S=1$, $B=Q=5$, $A=B=9$ respectively).
  • If $M=0$, we have the collision $M=U=0$.

In case Y, equation $(*)$ boils down to $10M- 4A+B =63$.

  • If $M\le5$, then $10M- 4A+B\le 50+B<63$; contradiction.
  • If $M=6$, we get $B=4A+3$; this leaves $(A.B)=(7,1)$ with the collision $B=S=1$ and the good solution $(A,B)=(0,3)$.
  • If $M=7$, we get $4A-B=7$. This leaves the cases $(A.B)=(2,1)$, $(3,5)$, $(4,9)$ which respectively collide with $B=S=1$, $B=N=5$, $A=Q=4$.
  • If $M=9$,we get $4A-B=27$. This leaves the cases $(A.B)=(7,1)$, $(8,5)$, $(9,9)$ which respectively collide with $B=S=1$, $B=N=5$, $A=B=9$.

Only a single branch in this analysis has led to a solution:
$R=8$ $E=2$ $S=1$; $(N,Q,U)=(5,4,9)$; $M=6$ and $(A.B)=(0,3)$.


Square has a value of 149082

Number has a value of 596328

The way I obtained my answer was, that in my python class I created a program to solver word puzzles, such as SEND+MORE==MONEY, and this was just another one of those. So I just ran my program with that puzzle.

Python3 code to solve the puzzle:

import re
import itertools

def solve(puzzle):
    words = re.findall('[A-Z]+', puzzle.upper())
    unique_characters = set(''.join(words))
    assert len(unique_characters) <= 10, 'Too many letters'
    first_letters = {word[0] for word in words}
    n = len(first_letters)
    sorted_characters = ''.join(first_letters) + \
        ''.join(unique_characters - first_letters)
    characters = tuple(ord(c) for c in sorted_characters)
    digits = tuple(ord(c) for c in '0123456789')
    zero = digits[0]
    for guess in itertools.permutations(digits, len(characters)):
        if zero not in guess[:n]:
            equation = puzzle.translate(dict(zip(characters, guess)))
            if eval(equation):
                return equation

if __name__ == '__main__':
    import sys
    for puzzle in sys.argv[1:]:
        solution = solve(puzzle)
        if solution:

This code is not the one I wrote, but does the same thing.


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