Divisibility

6! = 6 × 5 × 4 × 3 × 2 × 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!?

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

What digit must replace the star to make the number a multiple of 11?

Can you reconstruct the long division calculation from the 'skeleton'?

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.

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

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

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.

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). The question asks you to explain the trick.

Chocolate bars come in boxes of 5 or boxes of 12. How many boxes do you need to have exactly 2005 chocolate bars?