Puzzles

First, look at how many times each number will appear in the product.

Now split the odd and even numbers.

As all the powers in the first bracket are even, the first bracket is a square number.

Next, take a factor of two out of each number in the second bracket.

The only odd powers involved are now in the last bracket. Dividing by $50!$ would make each of these powers even, hence the overall number would be square.

For which numbers $m$ is it possible to find a number $n$ such that $$\frac{1!\times 2!\times ...\times m!}{n!}$$ is a square number?

No. If one of them were a square number, then its square root must end in 1 or 9 (as this is the only way to make the final digit a one). So the square root is of the form $10n\pm 1$.

If $10(10n^2\pm 2n)+1$ is of the form 111...1, then $10n^2\pm 2n$ is also of the form 111...1 (as it has just had the final 1 taken off). But $10n^2\pm 2n$ is even and 111...1 is odd, so this is not possible.

128

The result of the multiplication can be written as:

where $a$, $b$, $c$ and $d$ are integers. Expanding the brackets gives:

Next, if $(bd-ac{)}^2$ and $(bc+ad{)}^2$ are expanded, we get:

And so:

We have written the product as the sum of two integers.

For which integers $a$, $b$, $c$ and $d$ can the result $(a^2+b^2)(c^2+d^2)$ be written as the sum of two square integers in more than one way?

If $n^2$ has all odd digits then the units digit of $n$ must be odd. It can be checked that $n$ cannot be a one digit number (except 1 or 3 as given in the question) as the tens digit will be even.

Therefore $n$ can be written as $10A+B$ where $A$ is a positive integer and $B$ is an odd positive integer.

Now consider the tens digit of this.

$100A$ has no effect on this digit. The tens digit of $20AB$ will be the units digit of $2AB$ which will be even. The tens digit of $B^2$ is even (as checked above). Therefore the tens digit of $n^2$ is even.

Hence 1 and 9 are the only square numbers where all the digits are odd.

For which bases is this not true?

Let $T_a$ represent the $a$^th triangle number. This means that $T_a=\frac{1}{2}a(a+1)$.

Suppose that for some integer $n$, $n^2\le T_a<(n+1{)}^2$. This means that:

But for every positive integer $a\le a^2$, so:

$n$ and $a$ are both positive integers, so:

Now consider $T_{a+2}$:

We know that $a\ge n$ and $T_a\ge n^2$, so:

And so $T_{a+2}$ is not between $n^2$ and $(n+1{)}^2$. So if a triangle number $T_a$ is between $n^2$ and $(n+1{)}^2$ then the next but one triangle number $T_{a+2}$ cannot also be between $n^2$ and $(n+1{)}^2$. So there cannot be more than two triangle numbers between $n^2$ and $(n+1{)}^2$.

Given an integer $n$, how many triangle numbers are there between $n^2$ and $(n+1{)}^2$?

Let $a^2$ and $b^2$ be the two square numbers.

Prove that 3 times the sum of 3 squares is also the sum of 4 squares.

There are 64 1×1 squares, 49 2×2 squares, 36 3×3 squares, 25 4×4 squares, 16 5×5 squares, 9 6×6 squares, 4 7×7 squares and 1 8×8 square on a chessboard.

This can be shown by counting how many positions the top left corner of the square can sit on. For example, the top left corner of a 5×5 square can be in the first four rows and columns of the board (otherwise the square will go off the board) and 4×4=16.

64+49+36+25+16+9+4+1=204.

How many rectangles are there on a chessboard?