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Yahoo!

  • 2 years ago

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

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  1. Yahoo!
    • 2 years ago
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    @apoorvk @lgbasallote @yash2651995

  2. lgbasallote
    • 2 years ago
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    @ParthKohli :D

  3. apoorvk
    • 2 years ago
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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.

  4. RaphaelFilgueiras
    • 2 years ago
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    |dw:1339156636029:dw| but we have to account the two factors two any idea?

  5. Yahoo!
    • 2 years ago
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    its confusing

  6. Yahoo!
    • 2 years ago
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    @Ishaan94 @amistre64

  7. apoorvk
    • 2 years ago
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    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.

  8. Yahoo!
    • 2 years ago
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    so how will we do it I gave up! @phi

  9. Yahoo!
    • 2 years ago
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    @apoorvk

  10. phi
    • 2 years ago
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    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.

  11. Yahoo!
    • 2 years ago
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    I am also thinking! but no result

  12. FoolForMath
    • 2 years ago
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    Very easy actually.

  13. Yahoo!
    • 2 years ago
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    How?

  14. FoolForMath
    • 2 years ago
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    Do you know the total number of factors of \( 13260\) ?

  15. Yahoo!
    • 2 years ago
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    yes

  16. Yahoo!
    • 2 years ago
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    6 but 2 is repeating twice

  17. FoolForMath
    • 2 years ago
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    These are the only the prime factors.

  18. apoorvk
    • 2 years ago
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    'factors' are not the same as 'prime factors', @Yahoo!

  19. Yahoo!
    • 2 years ago
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    yup!!

  20. FoolForMath
    • 2 years ago
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    \(13260 = 2^2 \times 3 \times 5 \times 13\times 17 \) Total number of factors: \[= (2+1)\times (1+1)\times (1+1)\times (1+1)\times (1+1)=48\] Total number of pairs\( = \frac {48}2 = 24 \)

  21. Yahoo!
    • 2 years ago
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    @FoolForMath u r correct

  22. Yahoo!
    • 2 years ago
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    but @FoolForMath can u tell wat u have done!

  23. FoolForMath
    • 2 years ago
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    I have solved the problem.

  24. Yahoo!
    • 2 years ago
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    @FoolForMath dont joke i am asking abt method!

  25. FoolForMath
    • 2 years ago
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    http://mathworld.wolfram.com/DivisorFunction.html

  26. Yahoo!
    • 2 years ago
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    Can anyone do this in another way!

  27. eliassaab
    • 2 years ago
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    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}

  28. eliassaab
    • 2 years ago
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    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}}

  29. apoorvk
    • 2 years ago
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    @eliassaab Sir, but there must be a way to find the no. of pairs using permutations - how do we do that?

  30. eliassaab
    • 2 years ago
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    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.

  31. apoorvk
    • 2 years ago
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    Thanks a million Sir!!

  32. eliassaab
    • 2 years ago
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    yw

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