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Today I solve an interesting recreational math problem which took some time.

First I will briefly explain the situation of the problem. Look above figure.

(1) a,b,c,d,e,f,g,h,i,j are one of 1,2,3,4,5,6,7,8,9,10

(2) the sum of each row and columns are the same. (i.e., a+b+c+d = a+e+g, ...)

Find a,b,c,d,...

I tried to solve this by making equations but for me, that direction was too complicated.

Instead, I noticed that e and f should be big, and putting $x$ be the sum of each row or column and make 4x = 55 + a + b + c + d, hence $x=\frac{55}{4} + \frac{a+b+c+d}{4}$, and from the assumption I have $14\leq x \leq 22$. letting $e=10, f=9$ and solve $x = a+10+g = d+ 9 + j$ with many trials and errors I obtain $x=18$ and $a=2, g=6, d=8, j=1, b=5, c=3, h=4, i=7$.

Is there any simple way to solve this problem? Please let me know the simple strategy.

phy_math
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1 Answers1

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Some restrictions could be derived: if $$x = a+b+c+d = g+h+i+j,$$ then $$ 2x+e+f = 55, $$ so $$ 18\le x \le 26. $$

But:
$x=26$ doesn't work, since then $e+f=3$, so $e=1$ (if $e<f$), and $a+g=25$.
$x=25$ doesn't work, since then $e+f=5$, so $e\le 2$ (if $e<f$), and $a+g\ge 23$.
$x=24$ doesn't work, since then $e+f=7$, so $e\le 3$ (if $e<f$), and $a+g\ge 21$.
$x=23$ doesn't work, since then $e+f=9$, so $e\le 4$ (if $e<f$), and $a+g\ge 19$, which is possible for $e=4$, but isn't possible for corresponding $f=5$ and $d+j=18$.

Actually there are ten solutions (with sums $18$, $19$, $20$, $22$) (with $a<g$, $e<f$, $b<c$, $h<i$).
Therefore there is no unique short algorithm to determine "proper" solution.
There should be some "trial-and-error" part here, I guess.

Just few examples :

 x=18:
 1 4 7 6
 9     10
 8 3 5 2

x=18:
 4 1 7 6
 9     10
 5 3 8 2


x=19:
 1 5 6 7
 8     9
10 2 4 3


x=20:
 4 1 5 10
 7     8
 9 3 6 2


x=22:
 8 2 3 9
 4     7
10 1 5 6
Oleg567
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