$$ A \times B \times C \times D \times E \times F = 7 \times 10^7 $$
How can I find the number of ordered solutions for integers (I mean for integers $A,B,C,D,E,F$) so that they can satisfy the above equation?
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The factorization of $7 \times 10^7 $ is simply $2^7 \times 5^7 \times 7 $. Suppose for now we restrict the search for positive integer solutions only. For non-negative integers $x_i, y_i, z_i $, let $A = 2^{x_1} \cdot 5^{y_1} \cdot 7^{z_1}, B = 2^{x_2} \cdot 5^{y_2} \cdot 7^{z_2} , \ldots, F = 2^{x_6} \cdot 5^{y_6} \cdot 7^{z_6} $.
Then their product is $2^7 \times 5^7 \times 7^1 = 2^{\sum_{j=1}^6 x_j} \times 5^{\sum_{j=1}^6 x_j} \times 7^{\sum_{j=1}^6 x_j} $.
$ \Rightarrow \displaystyle \sum_{j=1}^6 x_j = 7, \sum_{j=1}^6 y_j = 7, \sum_{j=1}^6 z_j = 1 $.
By Stars and Bars, like TravisJ pointed out, we have $ {7+6 - 1 \choose 6-1} = {12 \choose 5} $ ordered solution for $<x_i>$ and $<y_i>$, and ${1+6 - 1 \choose 6-1} = 6 $ ordered solution for $<z_i>$.
Multiplying them together gives a total of $ {12 \choose 5} ^2 \times 6 = 3763584 $ ordered solution for positive integers $A$ to $F$.
If we drop the constraint for positive integers, then we could have:
Case 1: 6 Positive integers
Case 2: 4 positive integers, 2 negative integers
Case 3: 2 positive integers, 4 negative integers
Case 4: 6 Negative integers
Combining all these cases together yields a total of $3763584 \times \left [ {6 \choose 0} + {6 \choose 2} + {6 \choose 4} + {6 \choose 6} \right ] = 120434688 $ ordered solutions.
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1@Fermat Thank You. Hopefully it helps the OP. – May 13 '15 at 22:12
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Is that the final answer? – Sofia Dawes May 13 '15 at 22:43
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Yes if that's what youre looking for – May 13 '15 at 22:45
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This is essentially the same as: How many ways are there to express a number as the product of groups of three of its factors? except instead of needing 3 distinct numbers you need 6.
First, factor $7\times 10^{7}=2^{7}5^{7}7^{1}$. Then apply the idea there. Keep in mind that your stars and bars method has a $5$ bars instead of $2$ in the solution. I'll let you read the post to see why it is true, but the final answer will be $\binom{7+5}{5}\binom{7+5}{5}\binom{1+5}{5}$.
