Talk:Sub-unit squares

Why all the answers except 1 end with 36

We know that all perfect squares always end in 1,4,5,6,9, or an even number of zeroes. In addition, we know that for a number that ends in 1,4,9 its tens digit will always be even (2,4,6,8,0). If it ends with 6, its tens digit will be odd (1,3,5,7,9). If it ends with 5, its tens digit will be 2.

The initial perfect square cannot end in 2, 3, 7, or 8 (see above).

The initial perfect square cannot end with 0, because then we cannot subtract 1 from that zero.

If the initial perfect square ends in 9: This can only be the initial square, since it cannot be produced by subtracting 1 from each digit. Then the subtracted result ends in 8. But then this cannot be a square. So, the initial square cannot end in 9.

If the initial perfect square ends with 5, then its tens digit must be 2. So, the result after subtracting 1 from each digit ends in 14. But if a perfect square ends in 4, its tens digit must be even, and 1 is odd. So, the initial perfect square cannot end in 5.

If the initial perfect square ends in 4, then the resulting number after subtracting 1 from each digit ends in 3. But a perfect square cannot end with a 3. So, the initial perfect square cannot end with 4

Except for 1, if the initial perfect square ends with 1, then the number after subtracting 1 from each digit ends with 0. But all perfect squares ending with 0 have a tens digit of 0. So, the initial perfect square must end with 11. But if a perfect square ends with 1, its tens digit must always be even. So, the initial square cannot end with 1.

Therefore, by elimination, the initial perfect square must end in 6.

If the perfect square ends in 6, then when we subtract 1 from each digit, the resulting number will end in 5, so that number's tens digit must be 2 if the result is a perfect square. So, the initial number must end in 36. --Wherrera (talk) 18:43, 23 September 2022 (UTC)