Well here's how I got it:
We know the number has to be divisible by 1-10 with a remainder of 1. So 1 LESS than the number we're looking for is evenly divisible by them.
Also, we can ignore the divisibles of 1-5 since those factors are inherent in the factors of 6-10. For example, anything divisible by 8 is also divisible by 2 and 4, etc.
So we're looking for a number divisible by 6, 7, 8, 9, and 10....plus one.
Let's look at the factors of those numbers:
6: 2,
3
7: 7
8: 2,
2, 2
9: 3, 3
10: 2, 5
Now, I've color-coded the factors here that have pairs. To find the lest common multiple of 6-10, you take out one of each of these pairs. We're left with:
6: 2
7: 7
8: 2
9: 3, 3
10: 2, 5
or: 2, 7, 2, 3, 3, 2, 5.
Multiply these remaining factors together to get the least common multiple:
2x7x2x3x3x2x5 =
2520
Now we need to add that 1 back in there go give us the remainder of 1 in the riddle:
2521.
But 2521 is not divisible by 11. So what do we do? Add another
2520 and try that. And just keep adding
2520 to the number until we find one that divides evenly by 11.
Eventually we come across the answer:
25201
