Problem Solving

. What is the total number of integers between 100 and 200 that are divisible by 3?
(A) 33
(B) 32
(C) 31
(D) 30


What is an easier way to solve this
Ans A

This is the way I usually do it:

200-100= 100

100/3=33.33

Answer is 33

3 x 34 = 102 (First Integer b/w 100-200, divisible by 3)
3 x 35 =105
.
.
3 x 66 = 199(Last Integer b/w 100-200, divisible by 3)

Therefore, number of integers between 100-200 divisible by 3 equals
66 - 34 + 1 = 33

mindruna wrote:This is the way I usually do it:

200-100= 100

100/3=33.33

Answer is 33

.. U know what .. that need not always be true ...

.. consider the example of the interval 200 - 300 .. applying ur method we would arrive at the conclusion that 33 numbers are divisible by 3 .. but in reality in the interval 200 - 300 there are 34 numbers divisible 3 ..

.. it all depends if the first number of the series is divisible by 3 ..

.. just to be safe i would go with RM's method ...

You guys are totally right. Thanks for the correction!

thanks guys:)

The question asks how many integers btw 100-200. Thus the beginning and ending numbers aren't included so thus can't be counted.

recocollins wrote:The question asks how many integers btw 100-200. Thus the beginning and ending numbers aren't included so thus can't be counted.

The begining and ending numbers aren't divisible by 3 anyway...

Here is the way I look at these. If you're looking for multiples of 3 between 100 - 200, you can rephrase the question as how many integers are between 102 - 198 inclusive? So 198 - 102 = 96. 96/3 = 32 + 1 = 33. Remember you add 1 (which is representing a set of consecutive numbers) because the rephrased statement says [u]inclusive[/u].

Here is the way I look at these. If you're looking for multiples of 3 between 100 - 200, you can rephrase the question as how many integers are between 102 - 198 inclusive? So 198 - 102 = 96. 96/3 = 32 + 1 = 33. Remember you add 1 (which is representing a set of consecutive numbers) because the rephrased statement says [u]inclusive[/u].