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Resources tagged with Divisibility similar to Code to Zero:

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Broad Topics > Numbers and the Number System > Divisibility

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Code to Zero

Stage: 5 Challenge Level: Challenge Level:1

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.

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Multiplication Magic

Stage: 4 Challenge Level: Challenge Level:1

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). . . .

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Take Three from Five

Stage: 3 and 4 Challenge Level: Challenge Level:2 Challenge Level:2

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?

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Mod 3

Stage: 4 Challenge Level: Challenge Level:1

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

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Modulus Arithmetic and a Solution to Dirisibly Yours

Stage: 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.

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Big Powers

Stage: 3 and 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

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.

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What Numbers Can We Make Now?

Stage: 3 and 4 Challenge Level: Challenge Level:2 Challenge Level:2

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

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Power Mad!

Stage: 3 and 4 Challenge Level: Challenge Level:2 Challenge Level:2

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

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Knapsack

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

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

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Novemberish

Stage: 4 Challenge Level: Challenge Level:1

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.

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Prime AP

Stage: 4 Challenge Level: Challenge Level:1

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?

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N000ughty Thoughts

Stage: 4 Challenge Level: Challenge Level:1

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. . . .

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Sixational

Stage: 4 and 5 Challenge Level: Challenge Level:1

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. . . .

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Why 24?

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

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.

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Divisibility Tests

Stage: 3, 4 and 5

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

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The Knapsack Problem and Public Key Cryptography

Stage: 5

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.

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Dirisibly Yours

Stage: 5 Challenge Level: Challenge Level:1

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.

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Expenses

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

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

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Public Key Cryptography

Stage: 5

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.

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Squaresearch

Stage: 4 Challenge Level: Challenge Level:1

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

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LCM Sudoku

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

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

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Obviously?

Stage: 4 and 5 Challenge Level: Challenge Level:1

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

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Elevens

Stage: 5 Challenge Level: Challenge Level:1

Add powers of 3 and powers of 7 and get multiples of 11.

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Check Codes

Stage: 4 Challenge Level: Challenge Level:1

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. . . .

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Transposition Fix

Stage: 4 Challenge Level: Challenge Level:1

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. . . .

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Peaches Today, Peaches Tomorrow....

Stage: 3 and 4 Challenge Level: Challenge Level:2 Challenge Level:2

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?

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Odd Stones

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

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.

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Fac-finding

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

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.

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The Chinese Remainder Theorem

Stage: 4 and 5

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."

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396

Stage: 4 Challenge Level: Challenge Level:1

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.

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There's Always One Isn't There

Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

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.