Hi DValenz,
You are right that for it to be a divisible by both 3 and 4 it would need to be divisible by 12.
The way of thinking of this is of a consecutive set that starts with the first 3 digit number divisible by 12 i.e 108, and ends with the last 12 digit factor of 12 i.e. 996. Also every 12 units you will have another number that is divisible by 12.
As such we have an evenly spaced set with the following values:
a = 108
L = 996
d = 12
From this you can determine the total number of items in the set to be ((L - a )/d ) + 1 so in this case it would be :
((996 - 108)/12) + 1 = 75.
You must add one because this is inclusive of the first and last terms.
Hope this helps.
Mo
Factor if 12
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Unlimitedgmat
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Always include the answer choices.dvalenz wrote: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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As Mitch has demonstrated, the fastest approach is to first estimate the answer, and then check whether only one answer choice works.dvalenz wrote:How many 3 digit integers are divisible by 3 and 4?
As Mo has demonstrated, you can also apply the formula for an arithmetic sequences.
If you're not into memorizing formulas, you might also consider this approach:
We're looking for 3-digit multiples of 12. So, let's list a few: 108, 120, 132, . . . 984, 996
So, how many number are in that list?
To find out, let's rewrite each value as a multiple of 12.
108 = (9)(12)
120 = (10)(12)
132 = (11)(12)
.
.
.
984 = (82)(12)
996 = (83)(12)
At this point, we can see that we need only determine the number of integers from 9 to 83 inclusive. To do so, we'll apply an easy-to-remember rule:
The number of integers from x to y inclusive equals y - x + 1
So, the number of integers from 9 to 83 inclusive = 83 - 9 + 1 = 75
Answer = A
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sairakarim07
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It is nice to see such understanding.GMATGuruNY wrote:Always include the answer choices.dvalenz wrote: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.
