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by eaakbari » Thu Apr 22, 2010 8:16 pm
You have not given the options but the answer should be 14
Whether you think you can or can't, you're right.
- Henry Ford
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by tpr-becky » Thu Apr 22, 2010 8:46 pm
I agree the answer is 14 - if n^2 is divisible by 98 that means (n)(n) has all the same factors as 98 - the prime factors of 98 are (7)(7)(2) - since n is an integer that means that each n must contain one 7 and one 2 (the second two is left in the numerator) if n is (7)(2) then it is divisible by 14.
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by francoisph » Fri Apr 23, 2010 6:30 am
14

thks indeed for explanation Guys !
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by francoisph » Mon May 03, 2010 3:07 pm
Hi Guys,

do you have a technique to resolve quickly this issue please?

5/9 . 5/12 . 23/48 . 11/24 . 3/7
What is the positive difference between the largest and smallest of the fractions above?
A 1/12
B 5/36
C 1/4
D 1/3
E 7/18
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by ssuarezo » Tue May 04, 2010 8:47 am
tpr-becky wrote:I agree the answer is 14 - if n^2 is divisible by 98 that means (n)(n) has all the same factors as 98 - the prime factors of 98 are (7)(7)(2) - since n is an integer that means that each n must contain one 7 and one 2 (the second two is left in the numerator) if n is (7)(2) then it is divisible by 14.
Hi Becky:
I always have problems with this kind of questions 'cause there are so many properties that i dont know which to use, so I get wrong answers.

1rst- Why n must contain 7 and 2 just because it's an integer? wny not 7 and 7? why did u leave the other 7 behind?
2nd- n^2 could be any multiple of 98, right? so n has to share the same factors as 98, right? so why cant n be 98 itselft, as the largest integer. 14 should be the lowest integer.
3rd. Would it be ok if I do n = root(98x)? I know it takes me nowhere, but, is this correct?

I read some pages about integers, and many exercises, but I can't figure out what to use in which cases.

Thanks for your patience.
Silvia
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by tpr-becky » Tue May 04, 2010 9:37 am
To your first question if (n)(n) is divisible by 98 then that means that you need to be able to cancel all of the prime factors from the denomiator with factors in the numerator - the prime factors of 98 and 7, 7 and 2 - but becuase you have 2 n's on top each has to contain the same factors - thus n must be divisible by 7 and 2 - I didn't leave the second 7 out, it would be taken care of by the second n. With this there will be a 2 left uncancelled in the numerator but that doesn't affect divisibility. thus [(7)(2)]^2 is divisible by 98 becuase you can get rid of the denominator.

This question did not have answers but n must have a 7 and a 2 - so yes, n can be any multiple of 14 - but this question didn't have answers so I went with 14.

you can use the root you were talking about but you have to say that n=root(98) so that would mean n=7root2. Since n is a postive integer you have to also take out the 2 from under the root so that n =14.
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by ssuarezo » Tue May 04, 2010 9:59 am
tpr-becky wrote:To your first question if (n)(n) is divisible by 98 then that means that you need to be able to cancel all of the prime factors from the denomiator with factors in the numerator - the prime factors of 98 and 7, 7 and 2 - but becuase you have 2 n's on top each has to contain the same factors - thus n must be divisible by 7 and 2 - I didn't leave the second 7 out, it would be taken care of by the second n. With this there will be a 2 left uncancelled in the numerator but that doesn't affect divisibility. thus [(7)(2)]^2 is divisible by 98 becuase you can get rid of the denominator.

This question did not have answers but n must have a 7 and a 2 - so yes, n can be any multiple of 14 - but this question didn't have answers so I went with 14.

you can use the root you were talking about but you have to say that n=root(98) so that would mean n=7root2. Since n is a postive integer you have to also take out the 2 from under the root so that n =14.
Thanks Becky ...
Still the first paragraph of your answer is not fully clear to me (even if these are basic concepts), but, the second and third paragrapg make complete sense to me .. I'll go with the root method for this kind of problems
Thanks again !
Silvia.
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