Hint to Riddle #8: Remainders

8. What is the smallest positive integer that leaves a remainder of 1 when divided by 2, remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and so on up to a remainder of 9 when divided by 10?

Lets think about what numbers (N) would leave a remainder of 7 when divided by 8...

7, 15, 23, 31, 39, 47 and so on

It's pretty obvious at this stage that these numbers are all one less than the 8 times table. Which makes sense. If N were one higher it would be exactly divisible by 8. But it isn't so it leaves a remainder of 7.

What can we do with this observation? Well for a number N to leave a remainder of x-1 when divided by x we now know that N+1 must be a multiple of x. (x being 8 in the above example.)

In our question this must be true for x=2, 3, 4, ... ... 9, & 10

If you use this technique you may be able to arrive at an answer. But is it the lowest answer?

There is a tool here you can use to try different values of N

Note: no information is sent to me, the calculation is done entirely locally, on your computer.
Meaning N+1 is 2000

N+1 has the following factors:
2  3  4  5  6  7  8  9  10  All Factors 

Where next?
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