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How many positive three-digit integers are divisible by both [#permalink]

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16 May 2012, 12:04

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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?

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]

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26 Jun 2012, 10:09

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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.
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Re: How many positive three-digit integers are divisible by both [#permalink]

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01 Oct 2012, 06:31

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 )
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Re: How many positive three-digit integers are divisible by both [#permalink]

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01 Oct 2012, 09:32

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.
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Re: How many positive three-digit integers are divisible by both [#permalink]

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15 Dec 2014, 15:24

Hi Bunuel i have one question. so we need to check manually and find out the least and greatest numbers that is divisible by 3 and 4? like in this case 108 is the least number. so we have to test for each number from 100-108 is divisible by 3 and 4 or not? is this the only method?

thanks-

Bunuel wrote:

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.

Hope it helps.

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Re: How many positive three-digit integers are divisible by both [#permalink]

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24 Jan 2016, 17:31

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