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How many positive three-digit integers are divisible by both [#permalink]
16 May 2012, 11:04

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Difficulty:

15% (low)

Question Stats:

87% (02:22) correct
13% (00:52) wrong based on 89 sessions

How many positive three-digit integers are divisible by both 3 and 4?

A. 75 B. 128 C. 150 D. 225 E. 300

I know how to solve this one... but it takes me ages to find what would be the largest three digit number divisible by 12. Any tips or trick on how I can quickly get to that number?

Re: How many positive three-digit integers are divisible by both [#permalink]
16 May 2012, 11:18

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Expert's post

alexpavlos wrote:

How many positive three-digit integers are divisible by both 3 and 4?

A. 75 B. 128 C. 150 D. 225 E. 300

I know how to solve this one... but it takes me ages to find what would be the largest three digit number divisible by 12. Any tips or trick on how I can quickly get to that number?

Thanks!

A number to be divisible by both 3 and 4 should be divisible by the least common multiple of 3 and 4 so by 12.

How to find the largest three-digit multiple of 12: 1,000 is divisible by 4, so is 1,000-4=996, which is also divisible by 3, so 996 is the largest three-digit integer divisible by 12.

Re: How many positive three-digit integers are divisible by both [#permalink]
26 Jun 2012, 09:09

cant say that my method is good, but still...

first, look at answer choices. u can see that these choices range widely. now divide 999 by 12 and get 83. so, u need an answer choice that is at most 83.

only 75 is less than 83. so, A is the answer.

_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

Re: How many positive three-digit integers are divisible by both [#permalink]
01 Oct 2012, 05:31

Multiplying 3 by 4 we get the smallest no. that is divisible by both 3 as well as 4 ... Therefore any number that is divisible by 12 is also divisible by 3 and 4 ...

Our numbers are to begin from 100 and end at 999 ...

The first three digit no. that is divisible by 12 is 108 , and the last three digit no. is 996

Now we can set up an A.P. using 108 as our first number, 996 as our last number D= 12 ..

so we get 108 , 120 , 132 .........996 ...

The nth term is 996 and to calculate the value of n we use the following formula :

Tn = a + (n-1)d

Therefore 996 = 108 + (n-1) 12

996 - 108 = (n-1) 12

888/12 = n-1

74 = n-1

n = 75 .. ( A )

_________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

Re: How many positive three-digit integers are divisible by both [#permalink]
01 Oct 2012, 08:32

LalaB wrote:

cant say that my method is good, but still...

first, look at answer choices. u can see that these choices range widely. now divide 999 by 12 and get 83. so, u need an answer choice that is at most 83.

only 75 is less than 83. so, A is the answer.

It is good, because with the given list of choices, it works. With another choice below 83, it would have been another story.

_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.