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Counting Factors

Is there an efficient way to work out how many factors a large number has?


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.

Helen's Conjecture

Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?

Divisible Digits

Age 11 to 14 Short Challenge Level:

I wrote down a 4-digit number that was divisible by 3, 4, 5 and 6, but I spilt a cup of tea on it and can only see the first two digits.

The first two digits are 95 (in that order). What were the last two?

This problem is adapted from the World Mathematics Championships
You can find more short problems, arranged by curriculum topic, in our short problems collection.