I simple PS

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].