Factors and multiples

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?

Complete the following expressions so that each one gives a four digit number as the product of two two digit numbers and uses the digits 1 to 8 once and only once.

I added together the first 'n' positive integers and found that my answer was a 3 digit number in which all the digits were the same...

In how many ways can the number 1 000 000 be expressed as the product of three positive integers?

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?

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!?

When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .

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?

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