## Number Properties

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### Number Properties

by jsche229 » Tue Mar 18, 2014 11:28 am
Hey all-

I am solving problem 5 in the MGMAT number properties book (12 ed). The problem is as follows:

If J is divisible by 12 and 10, is J divisible by 24?

The answer choices are yes, no and can't be determined.

The answer I keep getting is yes due to the prime factorization however the answer key says I am 1 "2" short. The reason is because the "2" when factoring 10 is a duplicate.

Can anyone help me understand why it is a duplicate? Thanks!

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by Matt@VeritasPrep » Tue Mar 18, 2014 11:55 am
jsche229 wrote:Hey all-

I am solving problem 5 in the MGMAT number properties book (12 ed). The problem is as follows:

If J is divisible by 12 and 10, is J divisible by 24?

The answer choices are yes, no and can't be determined.

The answer I keep getting is yes due to the prime factorization however the answer key says I am 1 "2" short. The reason is because the "2" when factoring 10 is a duplicate.

Can anyone help me understand why it is a duplicate? Thanks!
Good question! This comes up a lot.

An easy way to "see" why this doesn't work is to consider the LCM of 12 and 10, which is 60. 60 divides by both 12 and 10, but it doesn't divide by 24. So dividing by both 12 and 10 is NOT enough to make a number divisible by 24.

A more abstract way to demonstrate this would be as follows.

If x is divisible by 12, then the prime factorization of x must contain 2 * 2 * 3 (at least two 2s and one 3).

If x is divisible by 10, then the prime factorization of x must contain 2 * 5 (at least one 2 and one 5).

If x is divisible by 12 and by 10, then we take everything we've got above: x is divisible by at least two 2s, at least one 3, and at least one 5. So x is divisible by 2 * 2 * 3 * 5, or 60.

Notice that the "extra" 2 from 10 doesn't matter: since x divides by 12, we know we already have at least one 2, so this simply overlaps with the 2 we find in 10.

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by jsche229 » Tue Mar 18, 2014 12:00 pm
That now makes sense. Would you apply the LCM method to solving problems or the abstract? Thanks!

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by ceilidh.erickson » Tue Mar 18, 2014 1:29 pm
Always think in terms of the LCM when you're given several pieces of divisibility information about the same variable. In this case, we were given two pieces of information about J. One way to frame the question is:

"Given that J is divisible by 12 and 10, what's the smallest number that J could be? It could be 60, and 60 isn't divisible by 24."

We wouldn't have to worry about that if the question were:
If X is divisible by 12 and Y is divisible by 10, is XY divisible by 24?

In this case, we're looking at 2 different variables, so there's no way that their factors could overlap. The product XY would have to be divisible by 2*2*3 and 2*5, so it would be divisible by 24.
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by Abhishek009 » Wed Mar 19, 2014 9:18 am
jsche229 wrote:Hey all-

I am solving problem 5 in the MGMAT number properties book (12 ed). The problem is as follows:

If J is divisible by 12 and 10, is J divisible by 24?

The answer choices are yes, no and can't be determined.

The answer I keep getting is yes due to the prime factorization however the answer key says I am 1 "2" short. The reason is because the "2" when factoring 10 is a duplicate.

Can anyone help me understand why it is a duplicate? Thanks!

Find the LCM of 10 and 12 , it comes out to be 60

Now all multiples of 60 are divisible by both 10 and 12

So , the multiples of 60 are - 60 , 120 , 180 , 240 , 300 ........

No divide the first few terms to check if 24 is divided without leaving any remainder or not.

60/24 = Not divisible

120/24 = Divisible

180/24 = Not divisible

So we cant be certain , coz some numbers are divisible while some are not...
Abhishek

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by Rich.C@EMPOWERgmat.com » Wed Mar 19, 2014 11:21 am
Hi jsche229,

When dealing with questions that involve an LCM, here are some Number Properties to keep in mind:

1) If one number divides evenly into the other, then the LCM is the larger number

The LCM of 4 and 12 is 12

2) If the two numbers are different EVEN numbers (and not divisible into one another), then the LCM is the product of the numbers DIVIDED BY 2

The LCM of 4 and 6 = (4)(6)/2 = 12

3) If the two numbers are different ODD numbers or one ODD and one EVEN (and not divisible into one another), then the LCM is the product of the numbers

The LCM of 5 and 7 = (5)(7) = 35
The LCM of 3 and 4 = (3)(4) = 12

You'll find these rules might come in handy on a few Quant questions on Test Day.

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Rich

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