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Yahoo!
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How many ways can the number 13260 be written as the product of two factors?
 2 years ago
 2 years ago
Yahoo! Group Title
How many ways can the number 13260 be written as the product of two factors?
 2 years ago
 2 years ago

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Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
@apoorvk @lgbasallote @yash2651995
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
@ParthKohli :D
 2 years ago

apoorvk Group TitleBest 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.
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
dw:1339156636029:dw but we have to account the two factors two any idea?
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
its confusing
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
@Ishaan94 @amistre64
 2 years ago

apoorvk Group TitleBest 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.
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
so how will we do it I gave up! @phi
 2 years ago

phi Group TitleBest 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.
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
I am also thinking! but no result
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
Very easy actually.
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
Do you know the total number of factors of \( 13260\) ?
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
6 but 2 is repeating twice
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
These are the only the prime factors.
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.0
'factors' are not the same as 'prime factors', @Yahoo!
 2 years ago

FoolForMath Group TitleBest 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\( = \frac {48}2 = 24 \)
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
@FoolForMath u r correct
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
but @FoolForMath can u tell wat u have done!
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
I have solved the problem.
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
@FoolForMath dont joke i am asking abt method!
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
http://mathworld.wolfram.com/DivisorFunction.html
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Can anyone do this in another way!
 2 years ago

eliassaab Group TitleBest 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}
 2 years ago

eliassaab Group TitleBest 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}}
 2 years ago

apoorvk Group TitleBest 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?
 2 years ago

eliassaab Group TitleBest 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.
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.0
Thanks a million Sir!!
 2 years ago
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