Sash143 wrote:

Bunuel wrote:

Official Solution:

What is the number of integers from 1 to 1000, inclusive that are not divisible by 11 or by 35?

A. 884

B. 890

C. 892

D. 910

E. 945

# of multiples of 11 in the given range \(\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(990-11)}{11}+1=90\);

# of multiples of 35 in the given range \(\frac{(\text{last}-\text{first})}{\text{multiple}}+1=\frac{(980-35)}{35}+1=28\);

# of multiples of both 11 and 35 is 2 (\(11*35=385\) and 770);

So, # of multiples of 11 or 35 in the given range is \(90+28-2=116\). Thus numbers which are not divisible by either of them is \(1000-116=884\).

Answer: A

Hi Bunuel,

I had come across a similar question "

How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?". In here we find # of multiples of 13 and 12, and eliminate multiples of both i.e 12*13 = 156. I understand since 156 has been counted both in # of 13's and # of 12's we delete 2 multiples from the answer.

But in the above problem both 11*35 = 385 and 2*11*35 = 770 must be counted twice in each of multiples of 11 and 13. So my doubt is why didn't we delete 4 multiples (385 and 770 are counted both in 11 multiples and 13 multiples) and instead did only 2 multiples (385 and 770).

Please help me understand!!!

There is one critical aspect of your question ("How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?") that is different from the question discussed in this forum.

If you can't spot it yet, its "but not both".

Your question asks us not to consider multiples of both 12 and 13.

The question discussed in this forum has no such restrictions. We are merely removing repetitions of multiples of both 11 and 35.

Before I sign off, here's the cliched "Hope it helps!"

So there are 90 multiples of 11 and 28 multiples of 35. You're absolutely correct in mentioning that there's 2 sets of repetitions for (385 and 770). However, if you subtract 4, you'll take both sets out. You only need to subtract 2 to still count the one set and eliminate the redundancy, not wipe them out completely.