Square numbers in squares

Peter placed the numbers 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 in the figure, so that each square contained exactly one number. Suddenly, his cat jumped on his table, knocking over his tea and six numbers disappeared (as shown in the figure).

In how many ways could he have placed the numbers, if by reading the numbers (from left) in each row we get a square number?


So for example if we place the 2 , 3 , 4 2, 3, 4 numbers in the third row in the figure above, then the 234 234 number would have to be a square number.

2 1 0 More than 4 3 4

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1 solution

Áron Bán-Szabó
Jun 11, 2017

a = v 2 \overline{a} = v^2

b 6 = x 2 \overline{b6} = x^2

c d e = y 2 \overline{cde} = y^2

f 0 g 5 = z 2 \overline{f0g5} = z^2

Since 5 z 2 5|z^2 , 5 z 5|z , but z z is not a multiply of 10 10 . So z z has 10 10 possible values: 5 , 15 , 25 , , 95 5, 15, 25, \dots, 95 . From these the 5 5 2 55^2 and 9 5 2 95^2 could be f 0 g 5 \overline{f0g5} .

If f 0 g 5 = 3025 \overline{f0g5}=3025 , then x x has only one possible value: x = 4 x=4 and b 6 = 16 \overline{b6}=16 . A square number's possible endings: 0 , 1 , 4 , 5 , 6 , 9 0, 1, 4, 5, 6, 9 , so a a and e e are 4 4 and 9 9 in a way. If a = 4 a=4 , then c d e \overline{cde} would be 789 789 or 879 879 , but non of these is a square number. So a = 9 a=9 and c d e \overline{cde} only could be 784 784 . So if f 0 g 5 = 3025 \overline{f0g5}=3025 , then there are one possible placing. If = 9025 \overline=9025 , then b 6 \overline{b6} only can be 36 36 , because if it would be 16 16 , then from the 3 , 4 , 7 , 8 3, 4, 7, 8 digits we would have to make two square numbers, but square numbers' endings are from the 0 , 1 , 4 , 5 , 6 , 9 0, 1, 4, 5, 6, 9 digits. So because neither 784 784 nor 874 874 is a square number, a = 1 a=1 , and because 874 874 is not a square number, c d e = 784 \overline{cde}=784 .

So there are two possible numbering.

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