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Find the number of integers divisible by 3...

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by BTGmoderatorLU » Tue Nov 21, 2017 7:52 am
Find the number of integers divisible by 3 in the interval from 10 to 200 inclusive.

(A) 61.
(B) 62.
(C) 63.
(D) 64.
(E) 65.

The OA is C.

I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
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by EconomistGMATTutor » Tue Nov 21, 2017 9:14 am
Hello LUANDATO.

We just need to find the first and the last multiple of 3 in that range, that is to say, the value of n such that $$10\le3\cdot n\le200.$$ By inspection we can conclude that $$4\le n\le66.$$ In this range there are $$\left(66-4\right)+1=63\ integers.$$ So, there are 63 numbers divisible by 3 in the given range,

The correct answer is C.

I really hope this can help you.

Feel free to ask me if you have a doubt.

Regards.
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by Brent@GMATPrepNow » Tue Nov 21, 2017 9:55 am
LUANDATO wrote:Find the number of integers divisible by 3 in the interval from 10 to 200 inclusive.

(A) 61.
(B) 62.
(C) 63.
(D) 64.
(E) 65.
A nice rule says: the number of integers from x to y inclusive equals y - x + 1
Example: the number of integers from 6 to 18 inclusive = 18 - 6 + 1 = 13

We can use the above rule to answer a wide range of questions.
First list a few values IN THE GIVEN RANGE:
12, 15, 18, 21, . . . . 195, 198

We can also write:
12 = 3(4)
15 = 3(5)
18 = 3(6)
.
.
.
195 = 3(65)
198 = 3(66)

So, the number of multiples is the same as the number of integers from 4 to 66 inclusive
The number of integers from 4 to 66 inclusive = 66 - 4 + 1 = 63
Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Sat Oct 12, 2019 4:06 pm
BTGmoderatorLU wrote:Find the number of integers divisible by 3 in the interval from 10 to 200 inclusive.

(A) 61.
(B) 62.
(C) 63.
(D) 64.
(E) 65.

The OA is C.

I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
The number of multiples of 3 is:

(198 - 12)/3 + 1 = 63

Answer: C

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