Problem:
How many positive integers $N$ less than or equal to $1000$ are there such that $75 \%$ of $N$'s divisors are multiples of $3$
$\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 37 \qquad \textbf{(E) } 38$
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 $\boxed{\textbf{(A)}}$.
Comments
Post a Comment