How many positive three-digit integers are divisible by both

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.

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 )

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)

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

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.
Cheers
RzS

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

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:

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

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

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-

Nice post. I hadn't seen it before.

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.

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.