# The number 4,000,000 has 63 positive integral factors. How do you find a and b, where 2^a 5^b is the product of all positive factors of 4,000,000?

Jun 21, 2016

$a = 8$ and $b = 6$

#### Explanation:

$4000000 = 4 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10$

= ${2}^{2} \times {10}^{6}$

= ${2}^{2} \times {\left(2 \times 5\right)}^{6}$

= ${2}^{2} \times {2}^{6} \times {5}^{6}$

= ${2}^{2 + 6} \times {5}^{6}$

= ${2}^{8} \times {5}^{6}$

Comparing it with ${2}^{a} {5}^{b}$, we get

$a = 8$ and $b = 6$

Jun 21, 2016

$a = 252$ and $b = 189$

#### Explanation:

$4000000 = 4 \cdot {10}^{6} = {2}^{8} \cdot {5}^{6}$

So each of the positive integral factors of $4000000$ is of the form ${2}^{m} \cdot {5}^{n}$ where $m \in \left\{0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8\right\}$ and $n \in \left\{0 , 1 , 2 , 3 , 4 , 5 , 6\right\}$.

Note that:

${\sum}_{m = 0}^{8} m = \frac{1}{2} 8 \left(8 + 1\right) = 36$

${\sum}_{n = 0}^{6} n = \frac{1}{2} 6 \left(6 + 1\right) = 21$

We use these below...

The product of all the factors is:

${\prod}_{m = 0}^{8} \left({\prod}_{n = 0}^{6} {2}^{m} \cdot {5}^{n}\right)$

$= {\prod}_{m = 0}^{8} \left({2}^{7 m} {\prod}_{n = 0}^{6} {5}^{n}\right)$

$= \left({\prod}_{m = 0}^{8} {2}^{7 m}\right) {\left({\prod}_{n = 0}^{6} {5}^{n}\right)}^{9}$

= 2^(7sum_(m=0)^8 m)*5^(9 sum_(n=0)^6 n

$= {2}^{7 \cdot 36} \cdot {5}^{9 \cdot 21}$

$= {2}^{252} \cdot {5}^{189}$

So $a = 252$ and $b = 189$