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

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

apoorvk
 3 years ago
Best ResponseYou've already chosen the best response.0find 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.

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

apoorvk
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, we can choose 1, 2,3 or 4 nos. at a time from this batch of 7 nos., in which a '2' is repeated.

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

phi
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

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

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

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

Yahoo!
 3 years ago
Best ResponseYou've already chosen the best response.06 but 2 is repeating twice

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

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

FoolForMath
 3 years ago
Best 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\( = \frac {48}2 = 24 \)

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

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

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

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

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

eliassaab
 3 years ago
Best ResponseYou've already chosen the best response.1As @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}

eliassaab
 3 years ago
Best ResponseYou've already chosen the best response.1Here 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}}

apoorvk
 3 years ago
Best 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?

eliassaab
 3 years ago
Best ResponseYou've already chosen the best response.1You 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.
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