anonymous
  • anonymous
How many ways can the number 13260 be written as the product of two factors?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@apoorvk @lgbasallote @yash2651995
lgbasallote
  • lgbasallote
@ParthKohli :D
apoorvk
  • apoorvk
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
|dw:1339156636029:dw| but we have to account the two factors two any idea?
anonymous
  • anonymous
its confusing
anonymous
  • anonymous
@Ishaan94 @amistre64
apoorvk
  • apoorvk
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.
anonymous
  • anonymous
so how will we do it I gave up! @phi
anonymous
  • anonymous
@apoorvk
phi
  • phi
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.
anonymous
  • anonymous
I am also thinking! but no result
anonymous
  • anonymous
Very easy actually.
anonymous
  • anonymous
How?
anonymous
  • anonymous
Do you know the total number of factors of \( 13260\) ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
6 but 2 is repeating twice
anonymous
  • anonymous
These are the only the prime factors.
apoorvk
  • apoorvk
'factors' are not the same as 'prime factors', @Yahoo!
anonymous
  • anonymous
yup!!
anonymous
  • anonymous
\(13260 = 2^2 \times 3 \times 5 \times 13\times 17 \) Total number of factors: \[= (2+1)\times (1+1)\times (1+1)\times (1+1)\times (1+1)=48\] Total number of pairs\( = \frac {48}2 = 24 \)
anonymous
  • anonymous
@FoolForMath u r correct
anonymous
  • anonymous
but @FoolForMath can u tell wat u have done!
anonymous
  • anonymous
I have solved the problem.
anonymous
  • anonymous
@FoolForMath dont joke i am asking abt method!
anonymous
  • anonymous
http://mathworld.wolfram.com/DivisorFunction.html
anonymous
  • anonymous
Can anyone do this in another way!
anonymous
  • anonymous
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}
anonymous
  • anonymous
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}}
apoorvk
  • apoorvk
@eliassaab Sir, but there must be a way to find the no. of pairs using permutations - how do we do that?
anonymous
  • anonymous
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.
apoorvk
  • apoorvk
Thanks a million Sir!!
anonymous
  • anonymous
yw

Looking for something else?

Not the answer you are looking for? Search for more explanations.