Question

Among the positive divisors of the number 12600, if $n_1$ is the number of divisors which are multiples of 3 and $n_2$ is the number of divisors which are multiples of 14, then $n_1+n_2=$

Answer 1

75

Answer 2

The number is 12600. First, we find the prime factorization of 12600.

$12600 = 126 \times 100$ $126 = 2 \times 63 = 2 \times 3^2 \times 7$ $100 = 10^2 = (2 \times 5)^2 = 2^2 \times 5^2$

So, $12600 = (2 \times 3^2 \times 7) \times (2^2 \times 5^2) = 2^{1+2} \times 3^2 \times 5^2 \times 7^1 = 2^3 \times 3^2 \times 5^2 \times 7^1$.

Any positive divisor of 12600 is of the form $d = 2^a \times 3^b \times 5^c \times 7^e$, where $a, b, c, e$ are integers satisfying the constraints:

$0 \le a \le 3$ $0 \le b \le 2$ $0 \le c \le 2$ $0 \le e \le 1$

We need to find $n_1$, the number of divisors which are multiples of 3. For a divisor to be a multiple of 3, the exponent of 3 in its prime factorization must be at least 1. So, $b \ge 1$.

The possible values for $a$ are $0, 1, 2, 3$ (4 options). The possible values for $b$ are $1, 2$ (2 options). The possible values for $c$ are $0, 1, 2$ (3 options). The possible values for $e$ are $0, 1$ (2 options). The number of such divisors is $n_1 = 4 \times 2 \times 3 \times 2 = 48$.

Next, we need to find $n_2$, the number of divisors which are multiples of 14. $14 = 2 \times 7$. For a divisor to be a multiple of 14, it must be divisible by both 2 and 7. This means the exponent of 2 must be at least 1 ($a \ge 1$) and the exponent of 7 must be at least 1 ($e \ge 1$).

The possible values for $a$ are $1, 2, 3$ (3 options). The possible values for $b$ are $0, 1, 2$ (3 options). The possible values for $c$ are $0, 1, 2$ (3 options). The possible values for $e$ are $1$ (1 option). The number of such divisors is $n_2 = 3 \times 3 \times 3 \times 1 = 27$.

Finally, we need to find $n_1 + n_2$. $n_1 + n_2 = 48 + 27 = 75$.