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Re: How many integers are divisible by 3 between 10! and 10! + 20 inclusiv [#permalink]

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16 Oct 2007, 09:55

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.

Re: How many integers are divisible by 3 between 10! and 10! + 20 inclusiv [#permalink]

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16 Oct 2007, 10:11

JDMBA wrote:

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!??????

Last edited by GMATBLACKBELT on 16 Oct 2007, 10:14, edited 1 time in total.

Re: How many integers are divisible by 3 between 10! and 10! + 20 inclusiv [#permalink]

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16 Oct 2007, 11:43

OlgaN wrote:

GMATBLACKBELT wrote:

yuefei wrote:

The question says "Inclusive"

Bah

Look, it is very simple. Try this: how many numbers from 1 to 100 inclusive? Not 100-1=99 NO NO NO It is 100-99+1=100. What is 1 here? It is the fist number in your question : 10!. You must count it if it is divisible by 3.

Do not be upset. I have known this only yesterday. My tutor explained it to me.

Thx, I get it, I just hate that I missed the "inclusive" part.

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 =(

Re: How many integers are divisible by 3 between 10! and 10! + 20 inclusiv [#permalink]

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25 Feb 2015, 09:53

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).

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

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

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

Re: How many integers are divisible by 3 between 10! and 10! + 20 inclusiv [#permalink]

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25 Dec 2015, 18:48

[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

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.

Re: How many integers are divisible by 3 between 10! and 10! + 20 inclusiv [#permalink]

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14 Mar 2016, 00:28

1

This post received KUDOS

GMATBLACKBELT wrote:

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

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.
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