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Yahoo!

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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  1. Yahoo!
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    @apoorvk @lgbasallote @yash2651995

    • one year ago
  2. lgbasallote
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    @ParthKohli :D

    • one year ago
  3. apoorvk
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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.

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

    • one year ago
  5. Yahoo!
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    its confusing

    • one year ago
  6. Yahoo!
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    @Ishaan94 @amistre64

    • one year ago
  7. apoorvk
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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.

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

    • one year ago
  9. Yahoo!
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    @apoorvk

    • one year ago
  10. phi
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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.

    • one year ago
  11. Yahoo!
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    I am also thinking! but no result

    • one year ago
  12. FoolForMath
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    Very easy actually.

    • one year ago
  13. Yahoo!
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    How?

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

    • one year ago
  15. Yahoo!
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    yes

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

    • one year ago
  17. FoolForMath
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    These are the only the prime factors.

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

    • one year ago
  19. Yahoo!
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    yup!!

    • one year ago
  20. FoolForMath
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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\( = \frac {48}2 = 24 \)

    • one year ago
  21. Yahoo!
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    @FoolForMath u r correct

    • one year ago
  22. Yahoo!
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    but @FoolForMath can u tell wat u have done!

    • one year ago
  23. FoolForMath
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    I have solved the problem.

    • one year ago
  24. Yahoo!
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    @FoolForMath dont joke i am asking abt method!

    • one year ago
  25. FoolForMath
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    http://mathworld.wolfram.com/DivisorFunction.html

    • one year ago
  26. Yahoo!
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    Can anyone do this in another way!

    • one year ago
  27. eliassaab
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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}

    • one year ago
  28. eliassaab
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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}}

    • one year ago
  29. apoorvk
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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?

    • one year ago
  30. eliassaab
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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.

    • one year ago
  31. apoorvk
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    Thanks a million Sir!!

    • one year ago
  32. eliassaab
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    yw

    • one year ago
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