What is the total number of integers between 100 and 200 that are divi

I go with A too

just (200-100)/3 = 33.33
so 33

yes there is a different way of arriving at that answer ....

u can also use airthmetic progression to get the answer

since the first term to be divisble by 3 is 102 ..take that as A .. the starting no

and since 198 is the last digit to be divisible by 3 take that as N ...

since the difference is 3 take that as D

no u have to find what term is 198 take that as nth term

the formula for that is N = A + (n-1) * d

198 = 102 +(n-1) * 3

from this u get n =33

though it seems that this way its takes longer to find the soultion ..but if u practise a bit of Arithemetic progression it will be a piece of cake ... and if u r learning Arithemetic progession , the definitely learn Geometric Progression too ...

Smallest number divisible by 3 = 102
Largest number divisible by 3 = 198

# of integers divisible by 3 = (198-102)/3 + 1 = 33

same method as ywilfred

step 1 smallest number div by 3 is 102
step 2 largerst num is 198
step 3 find the difference btw these 2 numbers = 198-102 = 96
step 4 div 96 by 3 and add 1 = 33

thus it is A

but ipc's approach is good...

In a set of consecutive integers, you'd expect about 1/3 of the numbers to be divisible by 3, since multiples of 3 are 3 apart, so A is the only reasonable answer here. There are many ways to do this rigorously. We can list all of the multiples of 3 in the given range - here it is useful to know that a number is divisible by 3 if the sum of its digits is divisible by 3. So 102 is the smallest multiple of 3 in the range, and 198 is the largest. We thus need to count how many numbers are in this list:

102, 105, 108, ..., 192, 195, 198

This is an equally spaced list; you can use the formula:

n = (largest - smallest) / ('space') + 1 = (198 - 102) / (3) + 1 = 96/3 + 1 = 32 + 1 = 33

Or if you don't want to use a formula, you could just divide everything in the list by 3:

34, 35, 36, ...., 64, 65, 66

then subtract 33 from each of these numbers:

1, 2, 3, ..., 31, 32, 33

to see that there must be 33 numbers in our original list.

Nice series of approaches to this question Ian!

1st number divisible by 3 between 100 and 200 is 102
Last number divisible by 3 between 100 and 200 is 198
Also, 102, 105, 108.... are the first three numbers divisible by 3 between 100 and 200.
common difference n = 3
Formula is t = a + (n-1)d
where t is the last number in the series
a is the first number in the series
n is the number of terms (we need to find this)
d is the common difference.
Substituting, 198 = 102 + (n-1) 3
solving, n = 33.
OA A

200-100 +1 =101/3=33.3

picked 33.

Note that this approach will not always work for such question.

For example:

What is the total number of integers between 102 and 202 that are divisible by 3?

Using the method as used above, you get 33.

But actually, the number of multiples of 3 here are 34.

102 = 3*34 (first multiple of 3 in the given sequence)
201 = 3*67 (last multiple of 3 in the given sequence)

All multiples from 34 to 67 are a part of this series. This gives us 67 - 34 + 1 = 34 multiples

Hey Karishma and Bunnel,

I wanted to have a reliable solution for such question and u provide me with that thanks alot.
Special thanks for Bunnel for formula.

Range = 100 ............ 200

First divisible by 3 = 102

Last divisible by 3 = 198

\(Count = \frac{198-102}{3} + 1 = 33\)

Answer = A

Total number of integers between 100 and 200 that are divisible by 3 = Total no of Interegers upto 200 divisible by 3 - Total no of Interegers upto 100 divisible by 3

Total no of Interegers upto 200 divisible by 3 = 200/3 =>66.67

Total no of Interegers upto 100 divisible by 3 = 100/ =>33.33

So,Total number of integers between 100 and 200 that are divisible by 3 = 66.67-33.33 =>33.34 ~ 33

Hence answer is (A) , 33

Using A(n)=A+(n-1)D
here N=33

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