How many numbers between 101 and 250, exclusive are divisible by three

Question

How many numbers between 101 and 250, exclusive are divisible by three?

A. 34
B. 41
C. 50
D. 83
E. 148

Abhishek009 · Accepted Answer

Numbers upto 250 divisible by 3 is 250/3 = 83
Numbers upto 101 divisible by 3 is 101/3 = 33

Total number of numbers between 101 and 250 that are divisible by 3 is 83 - 33 = 50

Hence correct answer will be (C) 50

stonecold · Answer

Digits are => 102,105,....249
Number of digits => 249-102/3 +1
=> 147/3 +1 = 49+1 => 50.
Wait a minute 50 isn't any of the options.

Am i missing something here  Bunuel

Regards
Stone Cold

Vinayak Shenoy · Answer

IMO C
numbers between 101 and 250, exclusive is given by {(249-102)/3}+1=49+1=50

Mended my answer. Thanks for the correction Austin.

mvictor · Answer

fastest way, find multiples of 3.
102 is a multiple of 3
252 is a multiple of 3, thus, 249 must be a multiple of 3.

102 is 34th multiple
249 is 83rd multiple.

83 - 34 = 49
since 102 and 249 are between 101 and 250, we need to add 1 (inclusive counting)
49+1 = 50.

answer is C.

ScottTargetTestPrep · Answer

Since we need to determine how many multiples of 3 are between 101 and 250, EXCLUSIVE, we can rewrite this as how many multiples of 3 are between 102 and 249, INCLUSIVE, and use the following equation:

(Largest multiple of 3 - smallest multiple of 3)/3 + 1

(249 - 102)/3 + 1 = 147/3 + 1 = 49 + 1 = 50

Answer: C

GMATinsight · Answer

Number that are divisible by 3 from 1 to 250 = 250/3 = 83
Number that are divisible by 3 from 1 to 100 = 100/3 = 33

i.e. Number that are divisible by 3 from 101 to 250 = 83-33 = 50

Answer: option C

rishit1080 · Answer

Got C.

Took me 1 minute and 23 seconds to just count it out the old fashion way.

stonecold · Answer

What if the question said between 101 and 999 :twisted:

broall · Answer

These numbers need to count have the common form: 3k Since 101 = 3*33 + 2, we have k > 33 Since 250 = 3*83 + 1, we have k <=83 So 34 <= k <= 83. => There are 83 - 34 + 1 = 50 numbers => OA is C

GMATinsight · Answer

Alternatively:

Solve it like an Arithmetic Progression (A.P.)

Smallest Number in the given range divisible by 3 = 102
Biggest Number in the given range divisible by 3 = 249

Let, 102 = first term
and 249 = nth term
Common difference, d = 3

nth Term = a+(n-2)d
i.e. 249 = 102+(n-1)*3
i.e. n = 50

Answer: Option C

rishit1080 · Answer

Then I would cry or try to learn this formula

GMATinsight · Answer

Then I would cry or try to learn this formula There is no formula to be learnt.. Please understand the basic concept as used in my first explanation in this thread. You can count multiples of any number x, from 1 to n by taking Integer solution of Calculate multiples of 3 from 1-250 and remove the multiples of 3 from 1-100 to find the number of multiples in the range 101-250 i.e. - = 83-33=50 I hope this helps!

rishit1080 · Answer

There is no formula to be learnt.. Please understand the basic concept as used in my first explanation in this thread.

You can count multiples of any number x, from 1 to n by taking Integer solution of

Calculate multiples of 3 from 1-250 and remove the multiples of 3 from 1-100 to find the number of multiples in the range 101-250

i.e.   -   = 83-33=50

I hope this helps!

So, for instance if it was multiples of 3 from 100-999 it would be

(999/3) - (100/3)= 333 - 33 = 300 ??

GMATinsight · Answer

So, for instance if it was multiples of 3 from 100-999 it would be (999/3) - (100/3)= 333 - 33 = 300 ?? Yes, absolutely correct.

Posted from my mobile device

diegocml · Answer

What an amazing approach to this question. So, when dividing 250/3 and 101/3, I just subtract their  quotients  while ignoring their  remainders , and the result is the number of integers within that range that are divisible by 3.

Abhishek009 · Answer

THere are numberous cool ways to solve this problem , including the AP form, however the method I have used to find is as below - Untitled.png Hope this helps ( Atleast to some extent ) :-D

bumpbot · Answer

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How many numbers between 101 and 250, exclusive are divisible by three

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