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

1. Yahoo! Group Title

@apoorvk @lgbasallote @yash2651995

2. lgbasallote Group Title

@ParthKohli :D

3. apoorvk Group Title

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 Group Title

|dw:1339156636029:dw| but we have to account the two factors two any idea?

5. Yahoo! Group Title

its confusing

6. Yahoo! Group Title

@Ishaan94 @amistre64

7. apoorvk Group Title

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! Group Title

so how will we do it I gave up! @phi

9. Yahoo! Group Title

@apoorvk

10. phi Group Title

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! Group Title

I am also thinking! but no result

12. FoolForMath Group Title

Very easy actually.

13. Yahoo! Group Title

How?

14. FoolForMath Group Title

Do you know the total number of factors of $$13260$$ ?

15. Yahoo! Group Title

yes

16. Yahoo! Group Title

6 but 2 is repeating twice

17. FoolForMath Group Title

These are the only the prime factors.

18. apoorvk Group Title

'factors' are not the same as 'prime factors', @Yahoo!

19. Yahoo! Group Title

yup!!

20. FoolForMath Group Title

$$13260 = 2^2 \times 3 \times 5 \times 13\times 17$$ 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! Group Title

@FoolForMath u r correct

22. Yahoo! Group Title

but @FoolForMath can u tell wat u have done!

23. FoolForMath Group Title

I have solved the problem.

24. Yahoo! Group Title

@FoolForMath dont joke i am asking abt method!

25. FoolForMath Group Title
26. Yahoo! Group Title

Can anyone do this in another way!

27. eliassaab Group Title

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 Group Title

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 Group Title

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

30. eliassaab Group Title

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 Group Title

Thanks a million Sir!!

32. eliassaab Group Title

yw