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An introduction to the ideas of public key cryptography using small numbers to explain the process. In practice the numbers used are too large to factorise in a reasonable time.
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. . . .
Take any pair of numbers, say 9 and 14. Take the larger number, fourteen, and count up in 14s. Then divide each of those values by the 9, and look at the remainders.
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?
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. . . .
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 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. . . .
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?
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."
Peter Zimmerman from Mill Hill County High School in Barnet, London gives a neat proof that: 5^(2n+1) + 11^(2n+1) + 17^(2n+1) is divisible by 33 for every non negative integer n.
You have worked out a secret code with a friend. Every letter in the alphabet can be represented by a binary value.
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). . . .
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.
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.
Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it.
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
Find and explain a short and neat proof that 5^(2n+1) + 11^(2n+1) + 17^(2n+1) is divisible by 33 for every non negative integer n.
Add powers of 3 and powers of 7 and get multiples of 11.
Find the values of n for which 1^n + 8^n - 3^n - 6^n is divisible by 6.
This article takes the reader through divisibility tests and how they work. An article to read with pencil and paper to hand.
An example of a simple Public Key code, called the Knapsack Code is described in this article, alongside some information on its origins. A knowledge of modular arithmetic is useful.
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?
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
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. . . .
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?
Take any prime number greater than 3 , square it and subtract one. Working on the building blocks will help you to explain what is special about your results.
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.
Find all 3 digit numbers such that by adding the first digit, the square of the second and the cube of the third you get the original number, for example 1 + 3^2 + 5^3 = 135.
On a "move" a stone is removed from two of the circles and placed in the third circle. Here are five of the ways that 27 stones could be distributed.