3 digit integers divisible by 3 and 4

[utkalnayak]

How many positive 3 integers are divisible by both 3 and 4 ?
A. 75
B. 128
C. 150
D. 225
E. 300

[GMATGuruNY]

How many positive three-digit integers are divisible by both 3 and 4 ?
A. 75
B. 128
C. 150
D. 225
E. 300

Since the answer choices here are far apart, we can BALLPARK.

The total number of 3-digit integers = 900. (From 100 to 999.)
To be divisible by 3 and 4, an integer must be a multiple of 12.
To ballpark how many multiples of 12 are contained within the 900 consecutive integers, simply calculate how many times 12 divides into 900:
900/12 = 75.

The correct answer is A.

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Hi utkalnayak,

Mitch's approach for this type of situation is right-on, but if you don't immediately "see" that approach, there's still 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

[Sanctus Real]

Hello Utkalnayak,

The responses you have received so far are spot on, I am only going to chip in my own recommendations.

Just check in the 3digits range given the multiple of 12 in the upper and lower limits.

For the upper limit you find that 996 is it, and the lower limit you find that 108 is it.

Hence, carry out your calculations thus...

(996 - 108)/ 12 = 74

This is close to the option 75 and hence our answer.

Hope this helps,

Cheers.

[ceilidh.erickson]

What is the source of this question? It seems highly unlikely that a real GMAT question would ask you to do this kind of heavy computation (or even complex estimation).

[GMATGuruNY]

What is the source of this question? It seems highly unlikely that a real GMAT question would ask you to do this kind of heavy computation (or even complex estimation).

The source is GMATPrep:



A similar problem appears in MGMAT's CATs:
https://www.manhattanprep.com/gmat/forum ... t2317.html

[GMATinsight]

How many positive 3 integers are divisible by both 3 and 4 ?
A. 75
B. 128
C. 150
D. 225
E. 300

To answer this question we must understand

Step 1) The number divisible by 3 as well 4 will be Divisible by 12

Step 2) The total Numbers Divisible by 12 in the range from 1 through 1000 can be calculated easily by dividing 1000 by 12
i.e. 1000/12 = 83 Numbers are divisible by 12 from 1 through 1000

Step 3) The total Numbers Divisible by 12 in the range from 1 through 100 can be calculated easily by dividing 100 by 12
i.e. 100/12 = 8 Numbers are divisible by 12 from 1 through 100

Step 4) Number between 100 to 1000 (all three digit numbers) that are divisible by 12 = 83-8 = 75

Answer: Option A

[GMATinsight]

However the best approach should be this

Step 1) The number divisible by 3 as well 4 will be Divisible by 12

Step 2) The total Numbers Divisible by 12 in the range from 1 through 1000 can be calculated easily by dividing 1000 by 12
i.e. 1000/12 = 83 Numbers are divisible by 12 from 1 through 1000

Since There will be lesser numbers divisible by 12 in the range 100 to 1000 (3 digit numbers) therefore the answer must be less than 83

Only option that satisfies is Option A

[Scott@TargetTestPrep]

How many positive 3 integers are divisible by both 3 and 4 ?
A. 75
B. 128
C. 150
D. 225
E. 300

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