Problem 1010

Solution:

Let $N$ be a 9-digit decimal number whose digits are all different, does not end with $5$, and does not contain a $0$.

First, since $N$ has 9 digits, and all digits are different and nonzero, the digits must be $1,2,3,4,5,6,7,8,9$ in some order.

Let us consider the sum of the digits:

$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$$

So $N$ is a permutation of $1$ through $9$, and its digit sum is $45$.

A square number modulo $9$ can only be $0,1,4,7$ (since the quadratic residues modulo $9$ are $0^2=0$, $1^2=1$, $2^2=4$, $3^2=0$, $4^2=7$, $5^2=7$, $6^2=0$, $7^2=4$, $8^2=1$).

But $N$ has digit sum $45$, so $N \equiv 0 \pmod{9}$.

Therefore, $N$ is divisible by $9$.

If $N$ is a perfect square, then its square root must also be divisible by $3$ (since $9$ is a square, and $N$ is divisible by $9$).

Let $N = k^2$, with $k$ divisible by $3$.

But $N$ does not end with $5$ or $0$. The possible last digits for a square are $0,1,4,5,6,9$.

But $N$ cannot end with $0$ or $5$ (by the problem statement), so the possible last digits are $1,4,6,9$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Now, consider the divisibility by $9$:

If $N$ is divisible by $9$, then $k$ must be divisible by $3$.

Let us check the possible endings for $k$ so that $k^2$ ends with $1,4,6,9$.

Squares ending with $1$:
$k$ ends with $1$ or $9$.

Squares ending with $4$:
$k$ ends with $2$ or $8$.

Squares ending with $6$:
$k$ ends with $4$ or $6$.

Squares ending with $9$:
$k$ ends with $3$ or $7$.

But since $N$ is a permutation of $1$ through $9$, it cannot end with $0$ (already excluded), and it cannot end with $5$.

Therefore, such a number $N$ cannot be a square.