ZdenekMickeCZ

This is a classic problem related to the Chinese Remainder Theorem. The number you're describing leaves a remainder of 1 when divided by 2, 2 when divided by 3, 3 when divided by 4, and so on. In other words, the number is 1 unit more than a multiple of 2, 2 units more than a multiple of 3, and so on. We can represent this system of equations: n ≡ 1 (mod 2) n ≡ 2 (mod 3) n ≡ 3 (mod 4) ... n ≡ 9 (mod 10) When considering the first two equations, we can search for a number that leaves a remainder of 1 when divided by 2 and 2 when divided by 3. This is 5. As we progress, we can build on this foundation. However, to find the smallest number that satisfies these conditions, you can take the least common multiple (LCM) of numbers from 1 to 10 and subtract 1. LCM(1,2,3,4,5,6,7,8,9,10) = 2520 So, the number = 2520 - 1 = 2519 Indeed, 2519 satisfies all the given conditions. When you divide 2519 by any number from 2 to 10, the remainder is always 1 less than the divisor.