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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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  1. Yahoo! Group Title
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    @apoorvk @lgbasallote @yash2651995

    • 2 years ago
  2. lgbasallote Group Title
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    @ParthKohli :D

    • 2 years ago
  3. apoorvk Group Title
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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.

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

    • 2 years ago
  5. Yahoo! Group Title
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    its confusing

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

    • 2 years ago
  7. apoorvk Group Title
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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.

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

    • 2 years ago
  9. Yahoo! Group Title
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    @apoorvk

    • 2 years ago
  10. phi Group Title
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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.

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

    • 2 years ago
  12. FoolForMath Group Title
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    Very easy actually.

    • 2 years ago
  13. Yahoo! Group Title
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    How?

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

    • 2 years ago
  15. Yahoo! Group Title
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    yes

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

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

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

    • 2 years ago
  19. Yahoo! Group Title
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    yup!!

    • 2 years ago
  20. FoolForMath Group Title
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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\( = \frac {48}2 = 24 \)

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

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

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

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

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

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

    • 2 years ago
  27. eliassaab Group Title
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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}

    • 2 years ago
  28. eliassaab Group Title
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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}}

    • 2 years ago
  29. apoorvk Group Title
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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?

    • 2 years ago
  30. eliassaab Group Title
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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.

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

    • 2 years ago
  32. eliassaab Group Title
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    yw

    • 2 years ago
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