M07-14

Hi Bunuel,
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Might be a very basic question.I took lot of time to solve this mainly i took time to find the highest mulitple of 35 < 1000.
is there a fast way to do this? How do we quickly find out 980 is the highest multiple of 35 < 1000,

Hi all, Bunuel,

What about the following solution?:
1000/11=90,xx so we have 90 multiples of 11 in 1000
1000/35=28,xx so we have 28 multiples of 35 in 1000
Now, because we counted a few multiples of both 11 and 35 we need to subtract them:
1000/(35*11)=2,xx so we need to subtract 2 multiples of both 11 and 35.

Altogether:
90+28-2=116 so we have 1000-116=884 integers that are NOT divisible by 11 or 35.

My true question is this: can we use this technique outside of numbers from [1 to X], with intervals: for example: "what is the number of integers between 754 and 1000 that are not divisible by 11 or 35?". You can't directly use the same technique and calculate 246/11, 246/35 etc. but I wonder if there is a way of adapting the above technique. Maybe by finding only one end of the range (say 990 for the number 11) instead of two (990 and 759 for the number 11)?

I hope this is not too confusing, but I feel there is a way of doing these exercises faster...

Thanks anyways!!

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.

Kindly seeking assistance

What is the difference between this question and M12-36 that asks about the number of positive integers less than 200 that are multiples of 13 or of 12 but not of both? In the solution of M12-36, it subtracts 156 (being the number of the common multiple of both 12 and 13), as follows:
# of multiples of 13....15-1
# of multiples of 12 ...16-1
Total 29.

So the common multiple was subtracted from both 15 and 16 whereas the two common multiples in question M07-14 were only subtracted once as in 90+28-2=116. How come it wasn't 90-2+28-2?

I hope you understand my question.

M12-36
How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?

So, here we don't want numbers which are both multiples of 13 and 12.
# of multiples of 13 is 15
# of multiples of 12 is 16
# of multiples of both 13 and 12 is 1. This one multiple (156) is counted both in 15 and 16, so to get the number of multiples of 13 or multiples of 12 but not both, we should subtract 1 from both sets: (15 - 1) + (16 - 1) = 29.

M07-14
Here we need numbers which are divisible by 11 or by 35. So, numbers which are divisible by both should be counted.

# of multiples of 11=90;

# of multiples of 35=28;

# of multiples of both 11 and 35 is 2. These two multiples are counted both in 90 and 28 but we want them to be counted only once, hence 90 + 28 - 2 = 116.

Hope it's clear.

Hi Bunuel

In a previous question in this set, I saw the approach where to find the number of multiples, we divided the total number by the given number and rounded it off.

in this case, # of multiple of 11 = 1001/11 or 91 (not 90)
Off course this is wrong answer.

can you suggest if i am applying the approach incorrectly or should we avoid using this approach all together?

Where did I use that approach?

The formula is below:
\(\# \ of \ multiples \ of \ x \ in \ the \ range = \)

\(=\frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).


How many multiples of 4 are there between 12 and 96, inclusive?

Last multiple of 4 IN the range is 96;
First multiple of 4 IN the range is 12;

\(\frac{96-12}{4}+1=22\).

OR: how many multiples of 5 are there between -7 and 35, not inclusive?

Last multiple of 5 IN the range is 30;
First multiple of 5 IN the range is -5;

\(\frac{30-(-5)}{5}+1=8\).

OR: How many multiples of 7 are there between -28 and -1, not inclusive?

Last multiple of 7 IN the range is -7;
First multiple of 7 IN the range is -21;

\(\frac{-7-(-21)}{7}+1=3\).

Can you please help me understand...
385 (11 x 35) and 780 (next multiple) is counted both in the multiples of 11 and multiples of 35. So should we not subtract these numbers twice?
So 90+28-4?

We need numbers which are divisible by 11 or by 35. So, numbers which are divisible by both should be counted.

# of multiples of 11=90;

# of multiples of 35=28;

# of multiples of both 11 and 35 is 2. These two multiples are counted both in 90 and 28 but we want them to be counted only once, hence 90 + 28 - 2 = 116.

Hope it's clear.

The 30-second method: You can use a little number sense and PS technique to crack this one in no time at all. Ask yourself, how many times will 11 fit into 100? 9 times. This will repeat for each set of 100, since the upper limit is set at 1000.

\(9*10=90\)

So far, then, we have 90 numbers to remove. Now, repeat the process for 35, but with a little fine-tuning. How many times will 35 fit into 100? 2 times. Into 200? 5 times, since the second hundred starts at 105. There cannot be more than three instances of 35 fitting into any given 100, since 3 * 35 = 105. For each set of 200, then, we should get about five 35s, and there are five sets of 200 in 1000. (We need not concern ourselves with the exact sequence of 2s and 3s per 200.)

\(5*5=25\)

There should be around 90 + 25, or 115 numbers to remove from consideration. Yes, there will be an overlap for each instance in which 11 and 35 cross paths, but even if 11 were 10, that would only happen twice out of our range of numbers.

\(1000-115=885\)

The answer must lie within 2 of 885, so (A), 884, is the only option that works. We can choose (A) with 100 percent confidence and spare ourselves the mental energy we may need for the next challenge.

- Andrew

Hi!
I don't understand why you substract only 2 numbers if you are counting 11*35 and 11*70 twice. Shouldn't we substract 4?---

Thank you!

Two numbers 11*35 and 11*70 are included in multiples of 11 as well as in multiples of 35. So, they are included TWICE. We need these two numbers to be included only ONCE. Thus we subtract 2.

Maybe a silly question...

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

Should not be something like "excluding"? When I interpreted this question I would look for the whole range.

Can you please clarify if I'm missing something?

Posted from my mobile device

Not sure I follow. We have integers from 1 to 1,000, inclusive, so the first 1,000 positive integers: 1, 2, 3, ..., 1,000 and want to find how many of them are not divisible by 11 or by 35. Why should it be "excluding" ?

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.