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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. _________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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)? _________________

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$$.

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

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

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.\\\

hope it will help you. _________________

kudos me if you like my post.

Attitude determine everything. all the best and God bless you.

Last edited by 321kumarsushant on 25 Apr 2012, 07:12, edited 3 times in total.

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.... _________________

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:

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