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Another n problem - divisibility

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Another n problem - divisibility

by gmat1011 » Sat Jun 26, 2010 6:26 am
I got this one pretty early on on a Kaplan mock GMAT today (I got the answer e by plugging in values and testing [spoiler]n=60[/spoiler] but it took away precious seconds - I could feel the seconds slipping away - you know that feeling... :x

How does one apply the factor foundation rule/prime boxes etc in this context to get the answer instead of plugging in numbers to test the result?

12=2*2*3; 10=2*5

If a number is divisible by 12 and 10, it is not necessarily divisible by which of the following?:

(a) 4
(b) 6
(c) 15
(d) 20
(e) 24
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by akdayal » Sat Jun 26, 2010 7:24 am
If a number is divisible by 12 and 10, it is not necessarily divisible by which of the following?:

(a) 4
(b) 6
(c) 15
(d) 20
(e) 24
In this problem We can say number is divisible by LCM (12 and 10). i.e number is multiple of 60
60 = LCM(12 1nd 10).

number will be 60k ( k = ....... , -1, 1, 2, ...)
clearly 60k will not be divisible by 24. Hence Ans (e)
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by selango » Sat Jun 26, 2010 7:37 am
n is divisible by both 12 and 10.

find the least common multiple of both numbers.

LCM(10,12)=60

Check 60 with the options.

60 is not divisible by 24.

Hence E
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by selango » Sat Jun 26, 2010 7:41 am
akdayal,

60 K is divisible by 24.for ex if k=2,120 is divisible by 24

60 k is not divisible by 24,if k is and odd integer.

60k/24=60k/12*2

5k/2[k must an even integer to make 60 divisble by 24]
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by akdayal » Sat Jun 26, 2010 8:07 am
@selango
5k/2[k must an even integer to make 60 divisble by 24]
Thanks for pointing me out.
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by neerajbansal » Sat Jun 26, 2010 8:37 am
This is how I looked at it.

If you think of 120 as one of the numbers to plug in ( multiple of 12 and 10 ) ....then you are remaining with 15 and 24 as the answer choices....

Then I took 60 and saw that 15 was out as a answer choice..

so i picked 24....



I like the LCM approach too...
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by tpr-becky » Sat Jun 26, 2010 8:43 am
If you think about this as factor division it works pretty quickly - you already said you did prime factorization for 12 and 10 to get a list of factors of 2, 2, 3, 2, 5 - that means you have those to work with But you have to remember that it only said they were divisible by 12 and 10 NOT 120 - so any factors that a duplicates across the numbers shoudl not be counted. This makes your new list 2,2,3,5

for something to be divisible then you have to have all of the same prime factors so you prime factor all of the answer choices and see which one has a factor that is not included in the above list.

A) 4 - 2,2
b) 6- 2, 3
c) 5,3
D) 2,2,5
E) 2,2,2,3 - -This one has too many 2's to work.
Becky
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The Princeton Review
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