How many integers are divisible by 3 between 10! and 10! + 20 inclusiv

Question

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

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

M25-02

Bunuel · Accepted Answer

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! is a multiple of 3 (10! = 2 *  3  * ... * 10), the question essentially asks how many integers divisible by 3 are there between a multiple of 3 and that multiple of 3 + 20, inclusive.
 
Thus, we can rephrase the question as: how many integers are divisible by 3 from 0 to 20, inclusive?
 
We can easily count 7 such integers: 0, 3, 6, 9, 12, 15, and 18.
 
Alternatively, use the formula to find the number of multiples of  x  in a range:  \frac{	ext{Last multiple of x in the range - First multiple of x in the range}}{x}+1 .
 
So,  \frac{18-0}{3}+1=7 .
 
Therefore, there are 7 integers divisible by 3 between  10!  and  10! + 20 , inclusive.

Answer: B

EMPOWERgmatRichC · Answer

Hi All,

This question is ultimately about "factoring" and why numbers divide evenly into other numbers.

I'm going to start with a simple example and work up to the details in this prompt:

You probably know that 3 divides evenly into 3! (3! = 1x2x3). We can factor out a 3 and get 3(2); mathematically, this means that 3 divides evenly into 3!

The same applies to 4! (4! = 1x2x3x4). We can factor out a 3 and get 3(1x2x4); so this means that 3 divides evenly into 4! In this same way, we know that 3 divides evenly into 5!, 6!, 7!, etc. We now know that 3 divides evenly into 10!.

Does 3 divide into 3! + 1? No, because you CAN'T factor out a 3. 
Does 3 divide into 3! + 2? No, because you CAN'T factor out a 3. 
Does 3 divide into 3! + 3? YES, because you CAN factor out a 3. You'd have 3(1x2 + 1).

This same rule applies to the range of values between 10! and 10! + 20

3 will divide evenly into: 
10! 
10! + 3 
10! + 6 
10! + 9 
10! + 12 
10! + 15 
10! + 18

Final Answer:  B

GMAT assassins aren't born, they're made, 
Rich

yuefei · Answer

Here's how I would approach this one:
(10! + 20) - 10! = 20 total integers

Since 10! doesn't include the integer 0, there are 20 integers possible. So 20 / 3 = 6 2/3 or 6 integers.

A. 6

GMATBLACKBELT · Answer

I said 6 too. This is a challenges problem, but I think the answer 7 is incorrect.

it says between 10! and 10! +20, so im guessin its a mistake.

JDMBA · Answer

B - 7

10! is divisible by 3 - The way I look Factorials is that any number included will also be divisible by the product. 10,9,8,7,6,5,4,3,2,1 are all divisors of 10!

There are 6 numbers between 10! and 10!+20 that are divisible by 3.

Hence 7

GMATBLACKBELT · Answer

Im just not getting this problem.

I know 10! is divisible by 3. U can just add up the digits of 10! and see that its divisible by 3. But...

it says the numbers between 10! and 10! +20, why are we including 10!??????

GMATBLACKBELT · Answer

Thx, I get it, I just hate that I missed the "inclusive" part. Like a problem the other day i did from MGMAT. https://www.gmatclub.com/forum/t53866 For some reason my brain was saying .9^2 is already multplied by itself, you dont have to do .9^2*.9^2. I hate it when I get like this, my mind refuses to look at the obvious =(

pacifist85 · Answer

There are a few things I don't understand with this question.

1) If 10! is 1x2x3x4x5x6x7x8x9x10, then why does it count only once (namely as a number 10!), as divisible by 3? So, I knew that 10! is divisible by 3, but I thought we needed to account for all the factors of 10! that are divisble by 3.

For example, 3,6,9 are divisible by 3. Also, 3*4=12, is also divisible by 3. Or 5x3=15 is also divisible by 3. This is why I was lost, because then there are numerous numbers that we can create that are divisible by 3.

But I guess the question clearly states that 10! is a number and I shouldn't have thought of it like 1x2x3x4x5x6x7x8x9x10. Right?

2) "There are 6 numbers between 10! and 10!+20 that are divisible by 3". Which numbers are those? How did you know what 10! is? Or you knew that it would end with 0, so the numbers that are divisible by 3 between 0 and 20 are 6 (3,6,9,12,15,18).

Thanx!

rohit8865 · Answer

Hi All,

This question is ultimately about "factoring" and why numbers divide evenly into other numbers.

I'm going to start with a simple example and work up to the details in this prompt:

You probably know that 3 divides evenly into 3! (3! = 1x2x3). We can factor out a 3 and get 3(2); mathematically, this means that 3 divides evenly into 3!

The same applies to 4! (4! = 1x2x3x4). We can factor out a 3 and get 3(1x2x4); so this means that 3 divides evenly into 4! In this same way, we know that 3 divides evenly into 5!, 6!, 7!, etc. We now know that 3 divides evenly into 10!.

Does 3 divide into 3! + 1? No, because you CAN'T factor out a 3.
Does 3 divide into 3! + 2? No, because you CAN'T factor out a 3.
Does 3 divide into 3! + 3? YES, because you CAN factor out a 3. You'd have 3(1x2 + 1).

This same rule applies to the range of values between 10! and 10! + 20

3 will divide evenly into:
10!
10! + 3
10! + 6
10! + 9
10! + 12
10! + 15
10! + 18

Final Answer:

Show Spoiler

B

GMAT assassins aren't born, they're made,
Rich

EMPOWERgmatRichC · Answer

Hi rohit8865,

Yes, some of the terms COULD end up factoring out 3^2, but we're not asked to do THAT math - we're just asked how many of the terms are divisible by 3. Each of the 7 numbers in the list are divisible by 3.

GMAT assassins aren't born, they're made,
Rich

stonecold · Answer

Nice Question
Here the rule i used is Multiple +Multiple = Multiple
hence 10!,10!+{3,6,9,12,15,18} are divisible by 3
hence B

SVaidyaraman · Answer

Since the question says inclusive , one has to first figure out whether 10! and 10!+20 are divisible by 3. We know that 10! is divisible by 3 and so 10! + 20 cannot be divisible by 3. Between them there are 6 numbers that are divisible by 3. So a total of 7.

BrentGMATPrepNow · Answer

There's a nice rule that says:   If M is divisible by k, and N is divisible by k, then (M + N) is divisible by k.   
Conversely,   If M is divisible by k, and Q is NOT divisible by k, then (M + Q) is NOT divisible by k.

First, since 10! = (10)(9)(8)..(  3  )(2)(1), we know that  10! is divisible by 3 .

So, by the   above rule  , we know that  10! + 3 is divisible by 3 
And  10! + 6 is divisible by 3 
 10! + 9 is divisible by 3 
 10! + 12 is divisible by 3 
 10! + 15 is divisible by 3 
 10! + 18 is divisible by 3

So, there are 7 integers from 10! to 10! + 20 inclusive that are divisible by 3.

Answer: B

Cheers,
Brent

manjot123 · Answer

since 10! is divisible by 3 between 10!---10!+20 there wil be 10!,10!+3...10!+18 total 7

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How many integers are divisible by 3 between 10! and 10! + 20 inclusiv

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How many integers are divisible by 3 between 10! and 10! + 20 inclusiv

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