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

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by [email protected] » Tue Jul 15, 2014 12:32 am
Hi shibsriz,

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

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by GMATGuruNY » Tue Jul 15, 2014 2:21 am
The intent of the problem seems to be as follows:
How many integers between 10! and 10!+20, inclusive, are divisible by 3?



6


7


8


9


10
Two number property rules:
(multiple of k) + (multiple of k) = (multiple of k).
(multiple of k) + (non-multiple of k) = (non-multiple of k).

Since 10! = 10*9*8*7*6*5*4*3*2, 10! is a multiple of 3.
To yield subsequent multiples of 3, we must add to 10! consecutive multiples of 3
Options between 10! and 10! + 20, inclusive:
10!
10! + 3
10! + 6
10! + 9
10! + 12
10! + 15
10! + 18.
Total options = 7.

The correct answer is B.
Last edited by GMATGuruNY on Fri Sep 05, 2014 11:36 pm, edited 1 time in total.
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by Brent@GMATPrepNow » Tue Jul 15, 2014 4:32 am
[email protected] wrote:How many integers are divisible by 3 between 10! and 10!+20 inclusive?
A) 6
B) 7
C) 8
D) 9
E) 10
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.

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

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by unknown13 » Wed Jul 16, 2014 12:54 am
hi
IMO the answer is B

following are the integer divisible by 3
10!
10! + 3
10! + 6
10! + 9
10! + 12
10! + 15
10! + 18
Total options = 7

thanks and regards
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by NandishSS » Mon Mar 05, 2018 7:07 am
GMATGuruNY wrote:The intent of the problem seems to be as follows:
How many integers between 10! and 10!+20, inclusive, are divisible by 3?

A)6
B)7
C)8
D)9
E)10
Two number property rules:
(multiple of k) + (multiple of k) = (multiple of k).
(multiple of k) + (non-multiple of k) = (non-multiple of k).

Since 10! = 10*9*8*7*6*5*4*3*2, 10! is a multiple of 3.
To yield subsequent multiples of 3, we must add to 10! consecutive multiples of 3?
Options between 10! and 10! + 20, inclusive:
10!
10! + 3
10! + 6
10! + 9
10! + 12
10! + 15
10! + 18.
Total options = 7.

The correct answer is B.
HI Guru/Brent,

If the question is How many integers between 10! and 10!+20, are divisible by 3?
A)6
B)7
C)8
D)9
E)10

What would be the ans?

Are you including 10!?

Thanks
Nandish
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by Jeff@TargetTestPrep » Tue Mar 06, 2018 10:36 am
[email protected] wrote:How many integers are divisible by 3 between 10! and 10!+20 inclusive?



6


7


8


9


10
Since 3 is a factor of 10!, 10! is divisible by 3. From this point, we can keep adding 3 to 10! to obtain integers that are divisible by 3, so the integers between 10! and 10!+20 inclusive that are divisible by 3 are:

10!, 10! + 3, 10! + 6, 10! + 9, 10! + 12, 10! + 15 and 10! + 18

We see that there are 7 such integers.

Answer: B

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