How many positive three-digit integers are divisible by both

Question

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

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

PS03344

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!

Bunuel · Accepted Answer

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.

abhasjha · Answer

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

LalaB · Answer

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.

vomhorizon · Answer

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 )

EvaJager · Answer

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

sunita123 · Answer

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-

pacifist85 · Answer

Nice post. I hadn't seen it before.

TheRzS · Answer

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  
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.
Cheers
RzS

warriorguy · Answer

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

A

EMPOWERgmatRichC · Answer

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....

Final Answer:  A

GMAT assassins aren't born, they're made, 
Rich

Sabby · Answer

Hi,

Three-digit numbers go from 100 to 999 included

How many three-digit numbers are there? 999-100 +1 = 899 + 1 = 900 three-digit numbers in total

Now, "divisible by 3 and 4" means divisible by 12 (3*4)

We divide the total number of three-digit by 12 : 900/12 = 75

Therefore, there are 75 three-digit numbers divisible by 4 and 3, or in other words by 12

Answer A)

HarishKumar · Answer

I know that between 1 and 100 there are 8 multiples of 12, ranging from 12 to 96.

I expect approximately similar numbers of multiples of 12 for every century block 201-299,301-399,401-499...901-999.

Since there are 9 blocks of 100 numbers each,

I know the answer would be close to 8 * 9 = 72.

The correct answer is  Choice A - 75 .

ScottTargetTestPrep · Answer

We need to determine how many numbers from 100 to 999 inclusive are divisible by 12.

Thus, we can use the formula of (last number in the set - first number in the set)/12 + 1

(996 - 108)/12 + 1

888/12 + 1

74 + 1 = 75

Answer: A

ArnauG · Answer

To determine the number of positive three-digit integers that are divisible by both 3 and 4, we need to find the count of integers that are divisible by the least common multiple (LCM) of 3 and 4, which is 12.

To find the count of three-digit integers divisible by 12, we need to determine the range of three-digit integers divisible by 12 and then calculate the count within that range.

The smallest three-digit integer divisible by 12 is 108 (9 * 12), and the largest is 996 (83 * 12).

To find the count, we need to calculate the number of terms in the arithmetic sequence formed by the multiples of 12 within this range.

We can use the arithmetic sequence formula: nth term = first term + (n - 1) * common difference.

The first term, a, is 108, the common difference, d, is 12, and the last term, l, is 996.

l = a + (n - 1) * d

996 = 108 + (n - 1) * 12

996 - 108 = (n - 1) * 12

888 = (n - 1) * 12

To solve for n, divide both sides of the equation by 12:

74 = n - 1

n = 75

So, the count of three-digit integers divisible by 12 is 75.

Therefore, the correct answer is (A) 75.

ImpactGMAT · Answer

3,4 LCM is 12

Now last 3 digit number divided by 12 is 996.

996/12 = 83

first 8 multiples of 12 are two digit numbers

83- 8 =75

ravi1522 · Answer

we can solve this by  series as number will be  in AP with difference of 12 (LCM of 3 &4)

difference=12
first number =108 
Last number =996

an=a1+ (n-1) * d

996 = 108 + (n-1) * 12

Solving, we get n=75

DanTheGMATMan · Answer

Looking for the multiples of 12 in the range:

https://www.youtube.com/watch?v=4W7GZ7Yil-E

NuttyIan · Answer

Realize that a multiple of 3 and 4 will only occur every 12 numbers. Then you realize there are 900 3 digit numbers, 100-999. Thus you divide by 900 and get 75, A.

egmat · Answer

Let me help you tackle this divisibility counting problem systematically.
Understanding the Core Requirement
When we need numbers divisible by  both  3 and 4, we're actually looking for numbers divisible by their LCM. Since 3 and 4 share no common factors, LCM(3,4) = 12.
So we're really counting:  How many three-digit multiples of 12 exist? 
Strategic Approach:
 Step 1: Find the boundaries

Smallest three-digit number: 100
 (\frac{100}{12} = 8.33...) 
So the smallest three-digit multiple is the 9th multiple:  (12 	imes 9 = 108)

Step 2: Find the upper boundary

Largest three-digit number: 999
 (\frac{999}{12} = 83.25) 
So the largest three-digit multiple is the 83rd multiple:  (12 	imes 83 = 996)

Step 3: Count the multiples 
We have multiples from the 9th to the 83rd, inclusive.
Using the counting formula:  (83 - 9 + 1 = 75)

This matches answer choice  A. 75

Master the pattern, not just this problem: See the full framework and practice similar problems here.

The  complete solution  reveals why the "+1" is essential here and demonstrates a verification technique that prevents common boundary errors. It also shows how this exact pattern appears in 3 other GMAT question types.

How many positive three-digit integers are divisible by both

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

How many positive three-digit integers are divisible by both

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards