2021 Fall OMC 10 #10

Problem:

How many positive integers $N$ less than or equal to $1000$ are there such that $75\mathrm{\%}$ of $N$'s divisors are multiples of $3$

$\mathbf{\text{(A) }}24\mathbf{\text{(B) }}25\mathbf{\text{(C) }}36\mathbf{\text{(D) }}37\mathbf{\text{(E) }}38$

Solution:

If ${\frac{3}{4}}^{\text{th}}$ of the divisors of $N$ are multiples of $3$, it means that ${\frac{1}{4}}^{\text{th}}$ of them are not. The ratio of these two fractions is $3$, which means that for every factor of $N$ which is not a multiple of $3$, there exist $3$ factors which are multiples of $3$. This implies that in the prime factorization of $N$, there exists the term $3^3$. Hence $N$ must be a multiple of $27$. The other condition is that there must be some other factor in the prime factorization of $N$ as well (the only prime number shouldn't be $3$). In the range $1\le N\le 1000$, the values of $N$ that satisfy the first condition are $$27\cdot 1,27\cdot 2,\dots 27\cdot 37,$$ which is $37$ numbers. Out of these, the ones that don't work are $$27\cdot 1,27\cdot 3,27\cdot 6,\dots ,27\cdot 36,$$ which is $13$ numbers. Hence the numbers that satisfy both conditions are $37-13=24$ numbers.

So, the answer is $\boxed{{\displaystyle \mathbf{\text{(A)}}}}$.