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How many ways can the number 13260 be written as the product of two factors?

Mathematics
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find out all the prime factors of the no. first..|dw:1339156205436:dw| So the prime factors are 2,2,3,5,17,13,1. Now for expressing this no. as a product of two factors, you need to find the no. of ways of choosing factors and pooling them into two separate groups, whose products would give the no. above, i.e. 13260.

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|dw:1339156636029:dw| but we have to account the two factors two any idea?
its confusing
Actually, we can choose 1, 2,3 or 4 nos. at a time from this batch of 7 nos., in which a '2' is repeated.
so how will we do it I gave up! @phi
  • phi
I would have to think about it, given enough time and perseverance I think I would get an answer. Why don't you work at it and post your (possible) solution.
I am also thinking! but no result
Very easy actually.
How?
Do you know the total number of factors of \( 13260\) ?
yes
6 but 2 is repeating twice
These are the only the prime factors.
'factors' are not the same as 'prime factors', @Yahoo!
yup!!
\(13260 = 2^2 \times 3 \times 5 \times 13\times 17 \) Total number of factors: \[= (2+1)\times (1+1)\times (1+1)\times (1+1)\times (1+1)=48\] Total number of pairs\( = \frac {48}2 = 24 \)
@FoolForMath u r correct
but @FoolForMath can u tell wat u have done!
I have solved the problem.
@FoolForMath dont joke i am asking abt method!
http://mathworld.wolfram.com/DivisorFunction.html
Can anyone do this in another way!
As @FoolForMath said, there are 48 divisors. Here they are: {1, 2, 3, 4, 5, 6, 10, 12, 13, 15, 17, 20, 26, 30, 34, 39, 51, 52, \ 60, 65, 68, 78, 85, 102, 130, 156, 170, 195, 204, 221, 255, 260, 340, \ 390, 442, 510, 663, 780, 884, 1020, 1105, 1326, 2210, 2652, 3315, \ 4420, 6630, 13260}
Here are the 24 pairs of factors {{1, 13260}, {2, 6630}, {3, 4420}, {4, 3315}, {5, 2652}, {6, 2210}, {10, 1326}, {12, 1105}, {13, 1020}, {15, 884}, {17, 780}, {20, 663}, {26, 510}, {30, 442}, {34, 390}, {39, 340}, {51, 260}, {52, 255}, {60, 221}, {65, 204}, {68, 195}, {78, 170}, {85, 156}, {102, 130}}
@eliassaab Sir, but there must be a way to find the no. of pairs using permutations - how do we do that?
You do not need permutation to do this. You take all the divisors in ascending order. You pair the first 1 and the last one. That the first solution. Now pair the second divisor with the one before last and so on. So the number of the solutions as @FoolOfMath said is half the number of divisors.
Thanks a million Sir!!
yw

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