Suppose an operator types a US Bank check code into a machine and transposes two adjacent digits will the machine pick up every error of this type? Does the same apply to ISBN numbers; will a machine. . . .

Details are given of how check codes are constructed (using modulus arithmetic for passports, bank accounts, credit cards, ISBN book numbers, and so on. A list of codes is given and you have to check. . . .

Find the values of n for which 1^n + 8^n - 3^n - 6^n is divisible by 6.

You have worked out a secret code with a friend. Every letter in the alphabet can be represented by a binary value.

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

Given any 3 digit number you can use the given digits and name another number which is divisible by 37 (e.g. given 628 you say 628371 is divisible by 37 because you know that 6+3 = 2+7 = 8+1 = 9). . . .

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.

What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time?

Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares?

I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number?

Imagine we have four bags containing numbers from a sequence. What numbers can we make now?

This article takes the reader through divisibility tests and how they work. An article to read with pencil and paper to hand.

What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2?

Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it.

Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.

6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?

Lyndon chose this as one of his favourite problems. It is accessible but needs some careful analysis of what is included and what is not. A systematic approach is really helpful.

The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?

Find the number which has 8 divisors, such that the product of the divisors is 331776.

How many numbers less than 1000 are NOT divisible by either: a) 2 or 5; or b) 2, 5 or 7?

The flow chart requires two numbers, M and N. Select several values for M and try to establish what the flow chart does.

What is the value of the digit A in the sum below: [3(230 + A)]^2 = 49280A

I put eggs into a basket in groups of 7 and noticed that I could easily have divided them into piles of 2, 3, 4, 5 or 6 and always have one left over. How many eggs were in the basket?

The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B?

Can you guarantee that, for any three numbers you choose, the product of their differences will always be an even number?

How many four digit square numbers are composed of even numerals? What four digit square numbers can be reversed and become the square of another number?

In this article for teachers, Bernard Bagnall describes how to find digital roots and suggests that they can be worth exploring when confronted by a sequence of numbers.

A number N is divisible by 10, 90, 98 and 882 but it is NOT divisible by 50 or 270 or 686 or 1764. It is also known that N is a factor of 9261000. What is N?

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. . . .

Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence?

Find the highest power of 11 that will divide into 1000! exactly.

Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.

Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5?

Show that if three prime numbers, all greater than 3, form an arithmetic progression then the common difference is divisible by 6. What if one of the terms is 3?

Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?

Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.

A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?

Mathematicians are always looking for efficient methods for solving problems. How efficient can you be?

Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for?

How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?

In this article we shall consider how to solve problems such as "Find all integers that leave a remainder of 1 when divided by 2, 3, and 5."

Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?

The four digits 5, 6, 7 and 8 are put at random in the spaces of the number : 3 _ 1 _ 4 _ 0 _ 9 2 Calculate the probability that the answer will be a multiple of 396.

Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.

The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .

A game that tests your understanding of remainders.

List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?

This package contains a collection of problems from the NRICH website that could be suitable for students who have a good understanding of Factors and Multiples and who feel ready to take on some. . . .

Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have?