next question: find the smallest integer n such that 840n is a perfect square
Originally posted by Divanhot:
Hi,
It is useful to list the members of each set, i.e.
E (i.e. the universal set) = {1, 2, 3, ..., 8}, A = {1, 2, 3, 4, 5}, B = {2, 3, 4}.
Then n(A union B) refers to the number of elements in the union of A and B,
and n(A intersect B)' refers to the number of elements in the complement of A intersect B (i.e. those elements that lie outside of the intersection).
Thanks!
Cheers,
Wen Shih
Originally posted by Divanhot:next question: find the smallest integer n such that 840n is a perfect square
Hi,
First, express 840 as a product of prime factors, i.e. (2^3)(5)(21).
Next, we must consider two facts:
1. Any perfect square must have powers which are divisible by 2.
2. 840n must have prime factors 2, 5 and 21.
Try it, thanks!
Cheers,
Wen Shih
Originally posted by wee_ws:Hi,
First, express 840 as a product of prime factors, i.e. (2^3)(5)(21).
Next, we must consider two facts:
1. Any perfect square must have powers which are divisible by 2.
2. 840n must have prime factors 2, 5 and 21.
Try it, thanks!
Cheers,
Wen Shih
Hi wen shih, 21 is not a prime factor ah......
so the accurate solution should be (2^3)(3)(5)(7) for 840 prime factors, and thus n= (2)(3)(5)(7)
Hi,
Thanks for pointing out my error :)
Cheers,
Wen Shih
welcome. :)
