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How many positive three-digit integers are divisible by both

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

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New post 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?

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

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New post 16 May 2012, 12:18
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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.

# of multiples of 12 between 100 and 999, inclusive is (last-first)/multiple+1=(996-108)/12+1=75 (check this: how-many-multiples-of-4-are-there-between-12-and-94862.html).

Answer: A.

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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New post 11 Apr 2014, 05:25
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Alternate solution :

Total multiple of 12 till 1000= 1000/12 = 83 .(concerned only about integral part)

Multiples of 12 till 100= 100/12 = 8

There fore multiple of 12 between 100 and 1000 = 83-8 = 75.

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

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New post 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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New post 01 Oct 2012, 06:31
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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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New post 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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New post 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.

# of multiples of 12 between 100 and 999, inclusive is (last-first)/multiple+1=(996-108)/12+1=75 (check this: how-many-multiples-of-4-are-there-between-12-and-94862.html).

Answer: A.

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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New post 05 Jan 2015, 08:55
Nice post. I hadn't seen it before.
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Re: How many positive three-digit integers are divisible by both  [#permalink]

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New post 22 Dec 2016, 16:45
Hi sunita123

Tad late to answer your question, but IMO YES you need to do it manually

But don’t worry, GMAT wont throw you some off the rail numbers. Lets consider 5 and 7
LCM=35

First 3 digit multiple of 35 = 70+35= 105 [simple mental math]
Last 3 digit multiple of 35 = 350+350+350 = 1050 …nope too much…subtract 70: 1050-70 = 980
Now (980-105)/35 + 1 = (875/35) + 1 = (700+175) + 1 = (35*20 + 35*5)/35 + 1 = 25+1 = 26

The takeaway is the mental math such as
- 875 is 700+175
and
- To find the last 3 digit multiple of 35, you get to 1050 first
etc.

Don’t know if it helps.
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Re: How many positive three-digit integers are divisible by both  [#permalink]

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New post 01 Jul 2017, 04:10
Another method is, the numbers form an A.P

difference=12

Last number =996

an=a1+ (n-1) * d

996 = 108 + (n-1) * 12

Solving, we get n=75

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

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New post 26 Feb 2018, 19:27
Hi All,

This question can be approached in a few different ways - and there's even a way to estimate the solution. You just have to do 'enough' work to spot the pattern.

We're looking for the number of 3-digit integers that are divisibly by BOTH 3 and 4.

Starting with the first 3-digit integer....

100 is divisibly by 4 but NOT 3

We can work "up" by adding 4 (since that would give us the "next" multiple of 4)....

104 is divisible by 4 but NOT 3

108 is divisible by 4 AND by 3

Notice the pattern so far...."miss", "miss", "hit"......

112 by 4 but NOT 3
116 by 4 but NOT 3
120 by 4 AND by 3

This also fits the pattern: "miss", "miss", "hit"....

It stands to reason that this pattern will continue, so we can leapfrog the misses and find the "hits" (notice that each is 12 greater than the prior one); here are the first several....

108, 120, 132, 144, 156, 168, 180, 192.....

So we have 8 multiples in the range of 100 - 200. Given this approximate pattern, there will probably be 8 or 9 terms in every set of 100 3-digit numbers. There are 9 groups of 100 from 100 to 999, so (approximately 8 per set)(9 sets) = about 72 multiples. There's only one answer that's close....

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

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Re: How many positive three-digit integers are divisible by both   [#permalink] 05 May 2019, 19:52
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