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# m25#02

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17 Oct 2008, 13:28
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How many integers are divisible by 3 between $$10!$$ and $$10! + 20$$ inclusive?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

[Reveal] Spoiler: OA
B

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Last edited by duuuma on 23 Oct 2008, 08:14, edited 1 time in total.
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28 Oct 2008, 08:39
Can someone show me the quick and easy way to do this?

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

6
7
8
9
10

IMO: 7
10!, 10!+3, 10!+6, 10!+9, 10!+12, 10!+15, 10!+18.
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28 Oct 2008, 11:42
i get 8... because when you divide (10!+9) you are still left with one more 3..from 9
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28 Oct 2008, 11:48
fresinha12 wrote:
i get 8... because when you divide (10!+9) you are still left with one more 3..from 9

if that is the case, then should you also consider 10!+18?
I will still stay with 7, though.
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28 Oct 2008, 12:18
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How can you have more than 7 when you're only talking about 21 numbers? 10! is the first, and 10!+20 is the 21st number. If the very first number is divisible by 3, then that means there are 7 numbers between 10! and 10!+20 that are divisible by 3.

10! is divisble by 3
10!+3 is too
10!+6 is too
10!+9 is too
10!+12 is too
10!+15 is too
10!+18 is too

that's 7

It's the same as saying "How many integers are divisible by 3 between 1! and 1!+20.

This question focuses more on the span between 10! and 10!+20 than the actual divisibility issue.

There will always be 7 when you have 21 consecutive numbers and determine how many are divisble by 3 because 21 is divisible by 3 seven times.
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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings SVP Joined: 29 Aug 2007 Posts: 2492 Followers: 69 Kudos [?]: 745 [0], given: 19 Re: m25 #2 [#permalink] ### Show Tags 28 Oct 2008, 12:38 jallenmorris wrote: How can you have more than 7 when you're only talking about 21 numbers? 10! is the first, and 10!+20 is the 21st number. If the very first number is divisible by 3, then that means there are 7 numbers between 10! and 10!+20 that are divisible by 3. 10! is divisble by 3 10!+3 is too 10!+6 is too 10!+9 is too 10!+12 is too 10!+15 is too 10!+18 is too that's 7 It's the same as saying "How many integers are divisible by 3 between 1! and 1!+20. This question focuses more on the span between 10! and 10!+20 than the actual divisibility issue. There will always be 7 when you have 21 consecutive numbers and determine how many are divisble by 3 because 21 is divisible by 3 seven times. jallenmorris, are you sure with the statement above marked red? Probably you want to say "How many integers are divisible by 3 between 0 and 20 (or 1!+20)? _________________ Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html GT SVP Joined: 30 Apr 2008 Posts: 1887 Location: Oklahoma City Schools: Hard Knocks Followers: 40 Kudos [?]: 578 [0], given: 32 Re: m25 #2 [#permalink] ### Show Tags 28 Oct 2008, 12:40 Isn't 1! = 1? and then 1! + 20 = 21? So how many integers are divisble between 1! and 1!+20 inclusive? (I didn't say inclusive before). The point of my statement is that the question does not mater that it's 10!, the key to the question is how many consecutive numbers there are in a row. Since there are 21 consecutive numbers, there will always be 7 integers in that set of 21 numbers that are divisible by 3. GMAT TIGER wrote: jallenmorris wrote: How can you have more than 7 when you're only talking about 21 numbers? 10! is the first, and 10!+20 is the 21st number. If the very first number is divisible by 3, then that means there are 7 numbers between 10! and 10!+20 that are divisible by 3. 10! is divisble by 3 10!+3 is too 10!+6 is too 10!+9 is too 10!+12 is too 10!+15 is too 10!+18 is too that's 7 It's the same as saying "How many integers are divisible by 3 between 1! and 1!+20. This question focuses more on the span between 10! and 10!+20 than the actual divisibility issue. There will always be 7 when you have 21 consecutive numbers and determine how many are divisble by 3 because 21 is divisible by 3 seven times. jallenmorris, are you sure with the statement above marked red? Probably you want to say "How many integers are divisible by 3 between 0 and 20 (or 1!+20)? _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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10 May 2010, 16:20
okay, what I really want to know is if there's a quicker way to figure out this problem. Afaik, you're not allowed to use a calculator on the official exam. So how did you figure out what 10! equals to in such a short time? Do you really have to literally write that all out on paper to figure out that number?

Do all this 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 on the paper? thats really time consuming and could easily make a mistake. I mean, yea sure, as soon as I can figure out what 10! equals to. I can definitely figure out whether it's divisible by 3 then on to the final step. But, figuring out what is 10! is really an issue here on a time-constraint basis
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12 May 2010, 00:26
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I'm not sure why you need to calculate the exact number of 10!. As far as this question is concerned, you just need to understand that 10! is divisible by 3. Please make sure you read through all posts of this thread carefully.

If you know what a factorial is, you know that 10! is surely divisible by all numbers less than and equal to 10. Use it as a fact. You don't have to calculate the exact 10!

I hope this makes sense. Sorry if I misunderstood your question.
Norlan wrote:
okay, what I really want to know is if there's a quicker way to figure out this problem. Afaik, you're not allowed to use a calculator on the official exam. So how did you figure out what 10! equals to in such a short time? Do you really have to literally write that all out on paper to figure out that number?

Do all this 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 on the paper? thats really time consuming and could easily make a mistake. I mean, yea sure, as soon as I can figure out what 10! equals to. I can definitely figure out whether it's divisible by 3 then on to the final step. But, figuring out what is 10! is really an issue here on a time-constraint basis

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29 Nov 2010, 07:01
my Ans is 7.
the Nos which are divisible by 3 between 10! and 10!+20 is
10!
10!+3
10!+6
10!+9
10!+12
10!+15
10!+18.
thats all.
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29 Nov 2010, 07:14
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Norlan wrote:
okay, what I really want to know is if there's a quicker way to figure out this problem. Afaik, you're not allowed to use a calculator on the official exam. So how did you figure out what 10! equals to in such a short time? Do you really have to literally write that all out on paper to figure out that number?

Do all this 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 on the paper? thats really time consuming and could easily make a mistake. I mean, yea sure, as soon as I can figure out what 10! equals to. I can definitely figure out whether it's divisible by 3 then on to the final step. But, figuring out what is 10! is really an issue here on a time-constraint basis

Hey Norlan,

i hope i can suggest you a better quick and easy way.

since we all know that 10!=10x9x8x7x6x5x4x3x2x1
there has to be no doubt about divisibility of 10! by 3 as it has 3 as a factor.
so, numbers will be divisible by 3 if it contains 3 as factor.
starting from first no:
10! - it is divisible by 3(it has 3 as factor)
10!+1- not divisible ; as 10! is divisible by 3 but not 1.
10!+2
10!+3--10!+3x1
10!+4
10!+5
10!+6--10!+3x2
10!+7
10!+8
10!+9--10!+3x3
10!+10
10!+11
10!+12--10!+3x4
10!+13
10!+14
10!+15--10!+3x5
10!+16
10!+17
10!+18--10!+3x6
10!+19
10!+20

only above colored nos have 3 as a factor.
so, only these Nos will be divisible by 3
if you will count these no. it will be equal to 7.\\\

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Last edited by 321kumarsushant on 25 Apr 2012, 06:12, edited 3 times in total.
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30 Nov 2010, 00:28
Counting the number of integers within a range
=> substract the extremes and add one

Counting the multiples of x within a range:
=> find the nearest multiples of x (to the extremes)
=> substract them and divide by x

Here:
(10!+18 -10!) / 3 = 6
6+1 = 7

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30 Nov 2010, 00:49
For me the approach was based on the fact that in consecutive numbers every third number will be divisible by 3...

so in any 21 consecutive numbers only 7 numbers can be divisible by 3, irerespective of from where the counting starts...so i simply ignored the 10! part....
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30 Nov 2010, 23:21
10! is divisible by 3.

So every third number after that will be as well. There are exactly 6 such numbers, 10!+3 ... 10!+18

So total numbers = 1+6 = 7
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12 Jan 2011, 07:35
Damn! completely forgot to include 10! in my list. I guess that's the kinda booby trap they're setting.
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22 May 2011, 03:17
Hi Friends,

This is an amazing forum .. I am preparing for GMAT and planning to take it by June end..

I just took the M25 free sectional test and got 29 correct out of 37 ... 3 very silly mistakes

How do you all think I do ???

Cheers!
Thanks
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22 May 2011, 05:48
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I think you're doing great! You must be very good at math .

Welcome to the forum. You should be able to find a lot in here. Let me know if you need extra guidance with our resources. Here's a link to a study plan for beginners just in case:

gmat-study-plan-for-gmat-novices-start-your-gmat-journey-80727.html

BearBelly wrote:
Hi Friends,

This is an amazing forum .. I am preparing for GMAT and planning to take it by June end..

I just took the M25 free sectional test and got 29 correct out of 37 ... 3 very silly mistakes

How do you all think I do ???

Cheers!
Thanks

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22 May 2011, 07:16
Hey dzyubam,

Thanks for the encouragement My undergrad subject was Engg. ...

and you won't believe what I am about to write:

This is me:

WE1: IT ... 2.5 yrs

WE2: Market Research and Analytics: 1 year till now....

How nd why did you make the switch .. How do u plan to project it in ur essays ???

dzyubam wrote:
I think you're doing great! You must be very good at math .

Welcome to the forum. You should be able to find a lot in here. Let me know if you need extra guidance with our resources. Here's a link to a study plan for beginners just in case:

BearBelly wrote:
Hi Friends,

This is an amazing forum .. I am preparing for GMAT and planning to take it by June end..

I just took the M25 free sectional test and got 29 correct out of 37 ... 3 very silly mistakes

How do you all think I do ???

Cheers!
Thanks

_________________

I was born intelligent but education ruined me !

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22 May 2011, 23:50
Hehe . Very funny. I'm not applying anywhere, I'm only helping out with this website.

Why I made the switch - I felt that Market Research was not my thing.

BearBelly wrote:
Hey dzyubam,

Thanks for the encouragement My undergrad subject was Engg. ...

and you won't believe what I am about to write:

This is me:

WE1: IT ... 2.5 yrs

WE2: Market Research and Analytics: 1 year till now....

How nd why did you make the switch .. How do u plan to project it in ur essays ???

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22 Oct 2011, 08:16
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There is no need to calculate 10! since on the other hand we have 10! + 20.

So for the difference we need how many are divisible by 3 ..so total no. of integers 20 + 1= 21

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Re: m25#02   [#permalink] 22 Oct 2011, 08:16

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