### N000ughty Thoughts

How many noughts are at the end of these giant numbers?

### Mod 3

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

### Novemberish

a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.

# Divisibility Tests

##### Stage: 3 and 4

Published January 1997,October 2010,February 2011.

Here is another method for testing divisibility by $7$ suggested by T.R.Mukundan

1. Subtract a multiple of $7$ from the number such that, on subtraction, the final (units) digit of the remainder is zero. (e.g. If the final digit of the given number is $7$, subtract $7$, if the final digit is $5$ subtract $35$). This is always possible since there is a unique number in the $7$ times table (up to $10 \times 7 = 70$) ending in each digit $0$ to $9$. (see note below)
2. Ignore the zero on subtraction and repeat the above procedure until only a number with two digits is obtained. If this number is divisible by $7$, then the original number is also divisible by $7$

#### Example:

Is $1778$ divisible by $7$?
The final digit of $1778$ is $8$, so we subtract $28$ ($4 \times 7$)
$$1778-28 = 1750$$
$1750$ is divisible by $7$ if and only if $175$ is, so we can ignore the $0$. The final digit of $175$ is $5$, so we subtract $35$ ($5 \times 7$)
$$175-35=140$$
$140$ is divisible by $7$ if and only if $14$ is, so we can ignore the $0$. But $14=2\times 7$, so $1778$ is divisble by $7$.

What are we actually saying here? $$1778= 4\times 7 + 10 \times 5 \times 7 + 100 \times 2 \times 7$$ $$1778=254 \times 7$$
So really this is a method for dividing by $7$, rather than just testing for divisibility. The divisibility test given in the article was rather hard to remember, so maybe in this case we would be better off just doing the division!

#### Note

In step 1 of the method, we used the fact that there is a unique number in the $7$ times table (up to $10 \times 7 = 70$) ending in each digit $0$ to $9$. Is this true in other times tables? It works for $3, 7,9, 11\ldots$. But it certainly doesn't work for $2$ - no multiple of $2$ ends in $1$. It turns out that works for any number not divisible by $2$ or $5$ (the prime factors of $10$). To understand why this is true, read this article on the Chinese Remainder Theorem.