In how many ways $1000000$ can be expressed as a product of $3$ integers?

Question

In how many ways $1000000$ can be expressed as a product of $3$ integers where each integer is greater than 1? Here $a \times b \times c$, $b \times a \times c...$ are considered to be same.

I've tried many ways. But couldn't solve anyway. please help me...

Answer 1

Given number is $2^6\times 5^6$. Any factor of it of the form $2^x\times 5^y$ with both $x,y$ not exceeding 6, possibly zero.

We want $$2^6\times 5^6= 2^{x_1}\times 5^{y_1}\times 2^{x_2}\times 5^{y_2}\times 2^{x_3}\times 5^{y_3}= 2^{x_1+x_2+x_3}\times 5^{y_1+y_2+y_3}$$

Now it is clear that we have to find $0\le x_i,y_i\le 6$ such that $x_1+x_2+x_3=6 $ and $y_1+y_2+y_3=6$. The number of solutions is the answer. This is not difficult as the problem is only about addition and that too with numbers upto 6.

Answer 2

I get $114$. Here's how:

Exponents on the three powers of $2$ can be (in ascending order):

all different

• $0,1,5$
• $0,2,4$
• $1,2,3$

two the same

• $0,0,6$
• $0,3,3$
• $1,1,4$

all three same

• $2,2,2$

and likewise for the three powers of $5$.

So for the "all different powers on the $2$s and all different powers on the $5$s" we have:

Example: pairing $0,1,5$ for $2$s with $0,1,5$ for $5$s:

• $(2^0 5^5) (2^1 5^0) (2^5 5^1)$
• $(2^0 5^0) (2^1 5^5) (2^4 5^1)$
• $(2^0 5^1) (2^1 5^0) (2^4 5^5)$
• etc...

But... as pointed out by @farruhota, the calculation includes cases where there is a factor of $1$ (i.e., $2^0 5^0)$ so we do not include those.

Go through all such cases and find $114$.

Here's a list:

{{2, 2, 250000}, {2, 4, 125000}, {2, 5, 100000}, {2, 8, 62500}, {2, 10, 50000}, {2, 16, 31250}, {2, 20, 25000}, {2, 25, 20000}, {2, 32, 15625}, {2, 40, 12500}, {2, 50, 10000}, {2, 80, 6250}, {2, 100, 5000}, {2, 125, 4000}, {2, 160, 3125}, {2, 200, 2500}, {2, 250, 2000}, {2, 400, 1250}, {2, 500, 1000}, {2, 625, 800}, {4, 4, 62500}, {4, 5, 50000}, {4, 8, 31250}, {4, 10, 25000}, {4, 16, 15625}, {4, 20, 12500}, {4, 25, 10000}, {4, 40, 6250}, {4, 50, 5000}, {4, 80, 3125}, {4, 100, 2500}, {4, 125, 2000}, {4, 200, 1250}, {4, 250, 1000}, {4, 400, 625}, {4, 500, 500}, {5, 5, 40000}, {5, 8, 25000}, {5, 10, 20000}, {5, 16, 12500}, {5, 20, 10000}, {5, 25, 8000}, {5, 32, 6250}, {5, 40, 5000}, {5, 50, 4000}, {5, 64, 3125}, {5, 80, 2500}, {5, 100, 2000}, {5, 125, 1600}, {5, 160, 1250}, {5, 200, 1000}, {5, 250, 800}, {5, 320, 625}, {5, 400, 500}, {8, 8, 15625}, {8, 10, 12500}, {8, 20, 6250}, {8, 25, 5000}, {8, 40, 3125}, {8, 50, 2500}, {8, 100, 1250}, {8, 125, 1000}, {8, 200, 625}, {8, 250, 500}, {10, 10, 10000}, {10, 16, 6250}, {10, 20, 5000}, {10, 25, 4000}, {10, 32, 3125}, {10, 40, 2500}, {10, 50, 2000}, {10, 80, 1250}, {10, 100, 1000}, {10, 125, 800}, {10, 160, 625}, {10, 200, 500}, {10, 250, 400}, {16, 20, 3125}, {16, 25, 2500}, {16, 50, 1250}, {16, 100, 625}, {16, 125, 500}, {16, 250, 250}, {20, 20, 2500}, {20, 25, 2000}, {20, 40, 1250}, {20, 50, 1000}, {20, 80, 625}, {20, 100, 500}, {20, 125, 400}, {20, 200, 250}, {25, 25, 1600}, {25, 32, 1250}, {25, 40, 1000}, {25, 50, 800}, {25, 64, 625}, {25, 80, 500}, {25, 100, 400}, {25, 125, 320}, {25, 160, 250}, {25, 200, 200}, {32, 50, 625}, {32, 125, 250}, {40, 40, 625}, {40, 50, 500}, {40, 100, 250}, {40, 125, 200}, {50, 50, 400}, {50, 80, 250}, {50, 100, 200}, {50, 125, 160}, {64, 125, 125}, {80, 100, 125}, {100, 100, 100}}

Answer 3

There are seven ascending partitions $(006)$–$(222)$ of $6$ into three $\geq0$ parts. We have to choose such a partition for the three exponents $x_1\leq x_2\leq x_3$ of $2$ and independently a partition for the three exponents $y_j$ of $5$. For each of the $49$ possible combinations we have to find out how many different factor triples we can form by permuting the entries of the chosen $y_j$-partition. If I have done this correctly I obtain the following (symmetric) table, whereby the rows are corresponding to the $(x_1x_2x_3)$ and the columns to the $(y_1y_2y_3)$:

$$\matrix{ &006&015&024&033&114&123&222\cr \cr 006:\quad &0&1&1&1&2&3&1\cr 015:\quad&1&4&4&2&3&6&1\cr 024:\quad&1&4&4&2&3&6&1\cr 033:\quad&1&2&2&1&2&3&1\cr 114:\quad&2&3&3&2&2&3&1\cr 123:\quad&3&6&6&3&3&6&1\cr 222:\quad&1&1&1&1&1&1&1\cr}$$ Adding all entries in this table gives $114$.