60. Find the smallest natural number that gives pairwise distinct remainders when divided by $2, 3, 4, 5, 6, 7$, $8, 9, 10$.
Solution. Let's examine all numbers that have this property. Denote the number we are considering with this property as $N$. If such a number is even, then its remainder when divided by 4 is 2. Then the remainder when divided by 6 is $4, \ldots$, the remainder when divided by 10 is 8. Then the remainder when divided by 3 is 1, the remainder when divided by 5 is 3, and so on. Thus, if $N$ is even, then $N+2$ is divisible by all numbers from 2 to 10, and thus $N \geqslant[2,3, \ldots, 10]-2$. Now let $N$ be odd. Similarly, we get that $N$ gives a remainder of 3 when divided by 4, a remainder of 5 when divided by 6, and so on. Further, $N$ gives a remainder of 2 when divided by 3 (since $N$ gives a remainder of 5 when divided by 6) and a remainder of 4 when divided by 5. The remainder when divided by 9 can only be 8 (since $N$ is not divisible by 3 and the odd remainders are occupied), and for the remainder when divided by 7, there are 2 options: 0 and 6. In the first case, $N+1$ is divisible by all numbers from 2 to 10, and thus $N \geqslant[2,3, \ldots, 10]-1$. In the second case, we get that $N+1$ is a multiple of $[2,3,4,5,6,8,9,10]=360$ and gives a remainder of 1 when divided by 7. The smallest such number is 1800, and thus $N=1799$ will be the answer.