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

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

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.

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

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!??????

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 =(

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!

[quote="EMPOWERgmatRichC"]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

but don't we r getting 2 3's in (10! + 9) as 3^2(1*2*4*5*2*7.....+1) ??

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

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

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.

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

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

Hi Bunuel,

I have kind of stupid question :D does between mean inclusive or not? (ex. between 1 and 10 is 1<x<10 or 1<=X<=10)

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