Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

How many integers are divisible by 3 between \(10!\) and \(10! + 20\) inclusive?

A. 6 B. 7 C. 8 D. 9 E. 10

Since 10! itself is a multiple of 3 (10!=2*3*...*10), then the question boils down to: how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?

Or: how many integers are divisible by 3 from 0 to 20, inclusive?

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

How many integers are divisible by 3 between \(10!\) and \(10! + 20\) inclusive?

A. 6 B. 7 C. 8 D. 9 E. 10

Since 10! itself is a multiple of 3 (10!=2*3*...*10), then the question boils down to: how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?

Or: how many integers are divisible by 3 from 0 to 20, inclusive?

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

So, \(\frac{18-0}{3}+1=7\).

Answer: B

Hi Bunnel, if there isn't the word "inclusive" do we still +1?

How many integers are divisible by 3 between \(10!\) and \(10! + 20\) inclusive?

A. 6 B. 7 C. 8 D. 9 E. 10

Since 10! itself is a multiple of 3 (10!=2*3*...*10), then the question boils down to: how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?

Or: how many integers are divisible by 3 from 0 to 20, inclusive?

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

So, \(\frac{18-0}{3}+1=7\).

Answer: B

Hi Bunnel, if there isn't the word "inclusive" do we still +1?

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

So, if it were: how many integers are divisible by 3 from 0 to 20, NOT inclusive, then the answer would be (18 - 3)/3 +1 = 6.

I didn´t pay attention to the actual range 10! to 10! + 20. I only knew that there were 21 positive integers.

I think that in any range of consecutive positive integers, 1 out of 3 of those consecutive integers will be divisible by 3 and the other 2 will not.

This can be explained algebraically in the following way: Suppose x is any positive integer. A range of 3 integers can be described as x + (x + 1) + (x + 2). This yelds 3x + 3, or 3(x + 1). Therefore this expression is divisble by 3.

If you have 21 integers, you have to do this 7 times.

Does this make sense?
_________________

Consider giving me Kudos if I helped, but don´t take them away if I didn´t!

I didn´t pay attention to the actual range 10! to 10! + 20. I only knew that there were 21 positive integers.

I think that in any range of consecutive positive integers, 1 out of 3 of those consecutive integers will be divisible by 3 and the other 2 will not.

This can be explained algebraically in the following way: Suppose x is any positive integer. A range of 3 integers can be described as x + (x + 1) + (x + 2). This yelds 3x + 3, or 3(x + 1). Therefore this expression is divisble by 3.

If you have 21 integers, you have to do this 7 times.

Does this make sense?

Yes. As in my solution: the question boils down to "how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?"
_________________

I didn´t pay attention to the actual range 10! to 10! + 20. I only knew that there were 21 positive integers.

I think that in any range of consecutive positive integers, 1 out of 3 of those consecutive integers will be divisible by 3 and the other 2 will not.

This can be explained algebraically in the following way: Suppose x is any positive integer. A range of 3 integers can be described as x + (x + 1) + (x + 2). This yelds 3x + 3, or 3(x + 1). Therefore this expression is divisble by 3.

If you have 21 integers, you have to do this 7 times.

Does this make sense?

Yes. As in my solution: the question boils down to "how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?"

Hi Bunuel,

Thanks for your answer. I understand your point, but I thought you were saying that the first integer of the 21 must be divisible by 3. Kind of like in a sequence a, b, c, a, b, c… in which a is divisible by 3 but b and c are not.

I was saying that if you have 21 consecutive positive integers, it really doesn´t matter if the first, second, or third is the one that is divisible by 3. If you have 21 consecutive positive integers, you will have 7 integers that are divisible by 3 anyway.

For instance, you have 7 integers divisble by 3 in all ranges 1-21, 2-22, 3-23, 4-24, etc...
_________________

Consider giving me Kudos if I helped, but don´t take them away if I didn´t!

I didn´t pay attention to the actual range 10! to 10! + 20. I only knew that there were 21 positive integers.

I think that in any range of consecutive positive integers, 1 out of 3 of those consecutive integers will be divisible by 3 and the other 2 will not.

This can be explained algebraically in the following way: Suppose x is any positive integer. A range of 3 integers can be described as x + (x + 1) + (x + 2). This yelds 3x + 3, or 3(x + 1). Therefore this expression is divisble by 3.

If you have 21 integers, you have to do this 7 times.

Does this make sense?

Yes. As in my solution: the question boils down to "how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?"

yes, but a certain thing to determine is whether the corner values of the range are divisible or not...

So the word "between" excludes the 10! from counting the multiples? Very tricky to understand... What if the question asked, "How many integers are divisible by 3 from and inclusive 10! and 10!+20 inclusive?

It then would be different, correct? Engr2012 _________________

Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.

So the word "between" excludes the 10! from counting the multiples? Very tricky to understand... What if the question asked, "How many integers are divisible by 3 from and inclusive 10! and 10!+20 inclusive?

How many integers are divisible by 3 between \(10!\) and \(10! + 20\) inclusive?

A. 6 B. 7 C. 8 D. 9 E. 10

Since 10! itself is a multiple of 3 (10!=2*3*...*10), then the question boils down to: how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?

Or: how many integers are divisible by 3 from 0 to 20, inclusive?

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

So, \(\frac{18-0}{3}+1=7\).

Answer: B

Hi Bunuel,

The question here says between 10! and 10 + 20! and that is why I did not count 10!. Shouldn't the question say from and then say inclusive?

Hello, I am struggling to understand why are we counting the "0" as multiple of 3. the range is from 0 to 20, but 3:0 is undefined, I thought the answer was 6.

Hello, I am struggling to understand why are we counting the "0" as multiple of 3. the range is from 0 to 20, but 3:0 is undefined, I thought the answer was 6.

0 is not a divisor of any integer, but a multiple of every integer. So, 0 is divisible by 3: 0/3 = 0 = integer.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

The range is between 10 ! and 10 ! +20 and factorial starts from 1 i.e 10 ! = 1*.....10 , so the first multiple of 3 is 3 ...why 0??? even though the range is from 1 to 20 ...?