Find the number of even factors of 126000

Question

What is an easy way to solve the problem? I can solve it by trying all possible even numbers, but I don't think it is a smart way.

Find the number of even factors of 126000.

Answer 1

$126000 = 2^4*3^2*5^3*7$

Every and only numbers of the form $2^i3^j5^k7^l$ where $0 \le i \le 4; 0 \le j\le2;0 \le k\le3 ;0\le l\le1$ are factors. To be an even factor $1 \le i \le 4$.

So there a $4$ possible powers of 2, $3$ possible powers of 3 ($0, 1,$ or $2$), $4$ possible powers of 5, and $2$ possible powers of 7. Thats $4*3*4*2 = 96$ even factors.

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or to see this clearly:

The only possible factors are: $(1|2|4|8|16)*(1|3|9)*(1|5|25|125)*(1|7)$. That's $5*3*4*2$ possible factors.

The even factors are: $(2|4|8|16)*(1|3|9)*(1|5|25|125)*(1|7)$. That's $4*3*4*2$ even factors.

Answer 2

$$ 126000 = 2 \cdot 63 \cdot 1000 = 2\cdot 9 \cdot 7 \cdot 2^3 \cdot 5^3 = 2^4 \cdot 3^2 \cdot 5^3 \cdot 7 $$ An even divisor has the form $$ 2^{p_1} \cdot 3^{p_2} \cdot 5^{p_3} \cdot 7^{p_4} $$ with $$ p = (p_1, p_2, p_3, p_4) \in \{ 1, \dotsc, 4 \} \times \{ 0,\dotsc, 2 \} \times \{ 0,\dotsc, 3 \} \times \{ 0,\dotsc, 1 \} $$ so there is at least one factor $2$ in its product.

There are $4 \cdot 3 \cdot 4 \cdot 2 = 96$ different choices for the powers $p$ and the same number of resulting even divisors.

Answer 3

An even factor of 126,000 is twice a factor of $63,000=(2^3)(3^2)(5^3)(7^1)$. Can you find the total number of factors id 63,000 from that prime factorization? That will count the even factors of $63,000·2=126,000$.