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

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Yahoo!Best ResponseYou've already chosen the best response.0
@apoorvk @lgbasallote @yash2651995
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
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.
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
dw:1339156636029:dw but we have to account the two factors two any idea?
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
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.
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
so how will we do it I gave up! @phi
 one year ago

phiBest ResponseYou've already chosen the best response.0
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.
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
I am also thinking! but no result
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.2
Very easy actually.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.2
Do you know the total number of factors of \( 13260\) ?
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
6 but 2 is repeating twice
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.2
These are the only the prime factors.
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
'factors' are not the same as 'prime factors', @Yahoo!
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.2
\(13260 = 2^2 \times 3 \times 5 \times 13\times 17 \) Total number of factors: \[= (2+1)\times (1+1)\times (1+1)\times (1+1)\times (1+1)=48\] Total number of pairs\( = \frac {48}2 = 24 \)
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
@FoolForMath u r correct
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
but @FoolForMath can u tell wat u have done!
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.2
I have solved the problem.
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
@FoolForMath dont joke i am asking abt method!
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.2
http://mathworld.wolfram.com/DivisorFunction.html
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Can anyone do this in another way!
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
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}
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
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}}
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
@eliassaab Sir, but there must be a way to find the no. of pairs using permutations  how do we do that?
 one year ago

eliassaabBest ResponseYou've already chosen the best response.1
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.
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
Thanks a million Sir!!
 one year ago
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