How many consecutive 0's are there

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by cramya » Sun Mar 15, 2009 9:18 am
Can you please tell how to apply the same with 12^k ?
12 = 2*3*3 = 3*4

4 would be the limting power, Ian please correct me if I am mistaken.

Lets say 15! u wanna find highest vlue of k for which 12^k is a factor

15/4 + 15/4^2(0)

3

12^3

U could test it with a calc.


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by goelmohit2002 » Sun Mar 15, 2009 9:46 am
Let's do the same with 20!.

We want to find highest value of k for which 12^k is a factor.

thus 20/4 + 20/4^2
= 5 + 1 = 6.

Thus k should be 6.

But if we try the same with calculator/excel, then the maximum value that comes for k is 8.

20! = 2432902008176640000
12^8 = 429981696
12^6 = 2985984

And both the above numbers evenly divide 20!.

There is something that probably we are missing. If it is not possible via the above method, then how to solve the same ?

Thanks
Mohit

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by DanaJ » Sun Mar 15, 2009 10:09 am
cramya wrote:
Can you please tell how to apply the same with 12^k ?
12 = 2*3*3 = 3*4

4 would be the limting power, Ian please correct me if I am mistaken.

Lets say 15! u wanna find highest vlue of k for which 12^k is a factor

15/4 + 15/4^2(0)

3
I tend to say no: remember, 4 is 2*2 and 2 is VERY common (actually, the most common) in factorials.
In 15! - you have:
3 - one 3
6 - one 3's
9 - two 3's
12 - one 3
15 - one 3

This makes for six 3's in 15!.

Now look at how many 2's you've got:
2 - one 2
4 - two 2's
6 - one 2
8 - three 2's
10 - one 2
12 - two 2's
14 - one 2

This makes for 2^11 or 2*2^10 or 2*4^5. You have 4 five times in 15!.
So 3 is INCORRECT, it's actually 5.
Last edited by DanaJ on Sun Mar 15, 2009 10:23 am, edited 1 time in total.

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by DanaJ » Sun Mar 15, 2009 10:20 am
goelmohit2002 wrote:Let's do the same with 20!.

We want to find highest value of k for which 12^k is a factor.

thus 20/4 + 20/4^2
= 5 + 1 = 6.

Thus k should be 6.

But if we try the same with calculator/excel, then the maximum value that comes for k is 8.

20! = 2432902008176640000
12^8 = 429981696
12^6 = 2985984

And both the above numbers evenly divide 20!.

There is something that probably we are missing. If it is not possible via the above method, then how to solve the same ?

Thanks
Mohit
In all my practice with the GMAT, the only type of questions I have encountered would be regarding the number of trailing zeros. Now, the problem you have presented seems much more difficult than this mainly because 12 is not the product of ONLY TWO prime numbers: it's actually the product of 2^2 * 3.

The reason why your calculations turned out the way the did is because 4 is not a prime number

The method outlined in the posts above (i.e. 144/5 - in this case 20/4) provides the multiples of 4. THE PROBLEM HERE stems from the fact that two numbers that are EVEN but NOT multiples of 4, when multiplied, provide another multiple of 4.
EXAMPLE: 6*10 = 60 = 15*4.

IMHO, you will NEVER encounter such a problem in the GMAT. They'll just stick to the trailing zeros, since it's far too complicated to ask for multiples of numbers that are the product of more than one even number. It's just TOO TIME-CONSUMING...

EDIT: the solution to your problem is to break 12 in prime factors and see what you've got:
1. with 3:
20/3 + 20/9 = 6 + 2 = 8 ---- 3^8

2. with 2:
20/2 + 20/4 + 20/8 + 20/16 = 10 + 5 + 2 + 1 = 18 ---- 2^18 = 4^9.

Pick the smallest between the powers of the two factors, which will be 8.

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by cramya » Sun Mar 15, 2009 10:59 am
Dana,
U r correct in taking 2's instead of 4's for 12^k. I made an error in dividing by 4 and was not sure so wanted Ian to check.

In all my practice with the GMAT, the only type of questions I have encountered would be regarding the number of trailing zeros

This may be true.
IMHO, you will NEVER encounter such a problem in the GMAT.

Difficult to say IMO.Its always good to know since I have seen questions atleast on this forum to find the highest power instead of trailing zeros.

Again opinions are subjective and everyone is entitled to have one...

Nice post and takeaways is what I see.... Good luck All!


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by goelmohit2002 » Sun Mar 15, 2009 11:53 am
Will this above method work for all prime no.s ?

For e.g. if we are required to find out max value of k in 144! for
a) 2^k
b) 3^k
c) 31^k

Will this method always work for all prime numbers ?

Thanks
Mohit

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by DanaJ » Sun Mar 15, 2009 11:54 am
cramya wrote: Difficult to say IMO.Its always good to know since I have seen questions atleast on this forum to find the highest power instead of trailing zeros.

Again opinions are subjective and everyone is entitled to have one...

Nice post and takeaways is what I see.... Good luck All!


Regards,
Cramya
Warning: might be OFFTOPIC...

Yeah, I guess you're right, cramya.... But as you've said, it's just my point of view. Here's my reasoning:

Say I were a content developer for ETS and my job would be to just sit in my pretty office writing new and fascinating math problems. The boss comes over and tells me that the question pool kindda lacks in problems about factorials. I start thinking: factorials... How to make that a bit more interesting?
You've probably noticed the wording of the problem: "number of trailing zeros", "number of consecutive zeros".... This is what I find interesting. As a smart (and possibly sadistic) content developer that I am <sighs>, I'd go for this type of problem instead of asking: "What's the highest power of 12 in 20!?", simply because I'm testing more than the test-taker's ability to find the highest power: I'm making him use his weapons of logic... The poor guy has to do some serious thinking before actually understanding what the problem is asking: trailing zeros --- power of 10 ---- power of 5.


Well, hope you enjoyed my story... And again, as you've mentioned, it's only a personal opinion... :D

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by cramya » Sun Mar 15, 2009 12:00 pm
Nice story or rather fairy tale for a Sunday afternoon. :P

If I worked in GMAC make no guesses or bets on who the next content developer I would be hiring is. Its DanaJ....no questions...

But fortunately neither do I work for GMAC nor will u be hired as a content developer atleast for the benefit of others. Jus kidding here....


Good luck Dana.

Regards,
CR

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by goelmohit2002 » Sun Mar 15, 2009 6:23 pm
goelmohit2002 wrote:Will this above method work for all prime no.s ?

For e.g. if we are required to find out max value of k in 144! for
a) 2^k
b) 3^k
c) 31^k

Will this method always work for all prime numbers ?

Thanks
Mohit
Experts please share your thoughts about the above....

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by Ian Stewart » Mon Mar 16, 2009 2:28 am
DanaJ wrote: IMHO, you will NEVER encounter such a problem in the GMAT. They'll just stick to the trailing zeros, since it's far too complicated to ask for multiples of numbers that are the product of more than one even number. It's just TOO TIME-CONSUMING...
I agree with Dana that you are not at all likely to see a question that asks something like "What is the largest positive integer k for which 12^k is a divisor of 200!", because so few test takers will be able to answer it. There are, however, at least four questions in GMATFocus (which contains the most recently retired questions available) which all test the same concept as the questions in this post, but which do not mention trailing zeros at all. These questions look something like this:

What is the largest integer k for which 7^k is a divisor of 70! ?

I've changed the numbers, but the underlying idea is the same. For such a question, if you know the quick method goelmohit outlined above, you can get the answer quickly ( 70/7 = 10, 70/49 ~ 1, 10+1 = 11). However, even if you don't know the method, it's still practical to get the answer within two minutes; it's not time consuming to write out all of the multiples of 7 up to 70 here, and prime factorize them, for example. Notice, however, that the question asks about a prime divisor (7), and not about a product of primes, which would be more complicated.
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by Ian Stewart » Mon Mar 16, 2009 2:45 am
goelmohit2002 wrote:
goelmohit2002 wrote:Will this above method work for all prime no.s ?

For e.g. if we are required to find out max value of k in 144! for
a) 2^k
b) 3^k
c) 31^k

Will this method always work for all prime numbers ?

Thanks
Mohit
Experts please share your thoughts about the above....
Dana gave an excellent explanation above of why you should *only* apply this method after breaking your number into primes, but yes, it will work for any prime number. So if we want to prime factorize say 20!, we can do that quickly enough:

20/2 = 10; 20/4 = 5; 20/8 ~ 2 (round down); 20/16 ~ 1 (round down)

so 2^18 is the largest power of 2 which divides 20!

Continuing for 3, and rounding down with each division:

20/3 ~ 6; 20/9 ~ 2

so 3^8 is the largest power of 3 which will divide 20!

We also have 20/5 = 4, and 20/7 ~ 2, and the remaining primes only divide 20! once, giving us:

20! = 2^(18) * 3^8 * 5^4 * 7^2 * 11 * 13 * 17 * 19

That's not the type of thing you'd normally need to do on the GMAT, but if you can prime factorize a number like 20! quickly, then questions on this topic should be fairly easy to answer.

Finally, since this thread seems to have attracted some interest, I'll offer three questions which are variations on the question in the original post.
You'd only see questions like the below at the 750-800 level of the test, and while I've never seen real GMAT questions similar to questions 2 or 3 below, they both can be solved very quickly once you see what to do.

1. What is the largest positive integer k for which 27^k is a divisor of 45! ?

2. If z is equal to the product of the first 12 positive even integers, what is the largest integer k for which 2^k is a factor of z?

3. What is the largest positive integer k for which 2^k is a divisor of 9! - 8! ?

Enjoy!
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by DanaJ » Mon Mar 16, 2009 3:15 am
I'm only going to write short answers for these babies.

1.[spoiler] [cubes are my faves] 7[/spoiler]

2. [spoiler]2 is not a factor of odd numbers, so it's the same as asking what is the power of 2^k in 24!. My answer: 22[/spoiler]

3. [spoiler]3, since 2^3 = 8 = 9 - 1[/spoiler]

Thanks Ian!

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by Ian Stewart » Mon Mar 16, 2009 4:28 am
Great work on the first two, Dana - I think you started question 3 in the right way, but perhaps didn't notice the factorial sign after the 8? I'll include the solution in spoiler tags:

[spoiler]
9! - 8! = 8!*(9-1) = 8*8! = (2^3)*8!

Now, using the methods from earlier in this thread, we can see that the largest power of 2 which divides 8! is 2^7, so (2^3)*8! will be divisible by 2^10.
[/spoiler]
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by DanaJ » Mon Mar 16, 2009 4:47 am
Yeah, you're right... I missed the 8!.

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by cramya » Mon Mar 16, 2009 4:02 pm
Thanks Ian for the nice probs.

Please post more challenging of any type if possible.

Regards,
Cramya