### N000ughty Thoughts

Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the number of noughts in 10 000! and 100 000! or even 1 000 000!

### Mod 3

Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.

### Common Divisor

Find the largest integer which divides every member of the following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.

# Number Rules - OK

##### Stage: 4 Challenge Level:

Solutions to parts, or all of this problem have been received from Kirsty, Gilland Aimee, Natasha, Hamish, Shona, Sheila and Alison Colvin all of Madras College, Maryof Birchwood Community High School, Warrington, Andrei of School 205, Bucharest, as well as Shu and Clement. Congratulations to all of you for some very good solutions. Jack of Madras College started to look at extensions to the pattern in part five of the question by considering cubes of numbers. I think there might be a problem here for a future edition of NRICH Jack!

### Part 1

Can you convince me that if a square number is multiplied by a square number the product is ALWAYS a square number?

Let the two square numbers be

$a^2$ and $b^2$

$a^2\times b^2 = a \times a \times b \times b =(ab)(ab) = (ab)^2$

Therefore if a square number is multiplied by a square number the product is ALWAYS a square number.

### Part 2

Can you convince me that if a square number is divided by a square number the product is ALWAYS a square number?

Let the two square numbers be

$a^2$ and $b^2$ where $a$ is greater than $b$

$\frac{a^2}{b^2} = \frac{a \times a}{b \times b} =\frac{a}{b} \times \frac{a}{b} = \left( \frac{a}{b} \right)^2$

Therefore if a square number is divided by a square number the quotient is ALWAYS a square number.

### Part 3

Can you convince me that no number terminating in $2,3,7$, or $8$ is a perfect square

A number ends in any of the digits $0-9$

When you square any number the final digit of the answer is formed from the square of the units digit of the original number.

Therefore the only possible endings for perfect squares are:

Ending in $1$:

$1^2$ or $9^2$

Ending in $4$:

$2^2$ or $8^2$

Ending in $5$:

$5^2$

Ending in $6$:

$4^2$ or $6^2$

Ending in $9$:

$3^2$ or $9^2$

So no number terminating in $2,3,7$, or $8$ is a perfect square.

### Part 4

Can you convince me that a number ending in $5$ cannot be a perfect square unless the digit in the ten's column is a $2$.

Consider the square of a number ending in $5$ and with the tens digits represented by the letter $b$.

The tens digit in the solution is formed from the units and tens digits of number being mulitplied, plus anything that is carried from the product of the units.

 $a$ .... .... $b$ $5$ $\times$ $a$ .... .... $b$ $5$ $5a$ .... .... $5b^2$ $5$ $5b$ $0$ $0$ $0$ $0$ $0$ $0$

The units digit of the product =

$5 + 0 + 0+ \dots +0$

The tens digit of the product is the remainder when $2 + 5b + 5b + 0 + 0 + \dots + 0$ is divided by $10$ (the rest is caried into the hundreds column). That is $2$ ($5b + 5b$ leaves a remainder of zero when divided by $10$ )

### Part 5

Convince me that the square of any number is equal to for example:

$10^2 = 9^2 + 9 + 10$

Let any number be $n$. I will need to show that: $$n^2 = (n-1)^2 + n-1 + n$$ $$n^2 = n(n-1)+n$$ $$\quad = (n-1)(n-1)+(n-1)+n$$ $$\quad = (n-1)^2+(n-1)+n$$ A useful representation of this result was offered by Kirsty, Gill and Amy