What is the smallest natural number $n$ for which $10^{n}-1$ is a multiple of 37?
(a) 6
(b) 5
(c) 4
(d) 3
(e) 2
The correct option is (d).

Observe that $10^{n}-1$ is a number that has all its digits equal to 9. Note, also, that a multiple of 37, of the form $37 \times n$, only ends in 9 if $n$ ends in 7. Therefore, the smallest multiples of 37 ending in 9 are $37 \times 7=259$, $37 \times 17=629$, and $37 \times 27=999$. Since $999=10^{3}-1$, it follows that $n=3$.