Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 28 Aug 2015, 02:42

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If d is a positive integer and f is the product of the first

Author Message
TAGS:
Intern
Joined: 09 Sep 2005
Posts: 21
Followers: 0

Kudos [?]: 2 [0], given: 0

If d is a positive integer and f is the product of the first [#permalink]  10 Sep 2005, 11:58
2
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

48% (02:13) correct 52% (01:05) wrong based on 165 sessions
If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?

(1) 10^d is a factor of f
(2) d>6 Not Sufficient.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-d-is-a-positive-integer-and-f-is-the-product-of-the-first-126692.html
[Reveal] Spoiler: OA
 Kaplan Promo Code Knewton GMAT Discount Codes GMAT Pill GMAT Discount Codes
Math Expert
Joined: 02 Sep 2009
Posts: 29100
Followers: 4721

Kudos [?]: 49610 [4] , given: 7400

Re: Number Properties from GMATPrep [#permalink]  05 Oct 2009, 04:58
4
KUDOS
Expert's post
DenisSh wrote:
Bunuel wrote:
2. Finding the number of powers of a prime number k, in the n!.
What is the power of 3 in 35!...

In the same way as for 5? i.e., 35/3 + 35/9 + 35/27 = 11 + 3 + 1 = 15.

Am I right?

Absolutely, here is the way to calculate the number of powers of a prime number k, in n!.
The formula is:
$$\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}$$ ... till $$n>k^x$$

What is the power of 2 in 25!
$$\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22$$

There is another formula finding powers of non prime in n!, but think it's not needed for GMAT.
_________________

Last edited by bb on 21 Oct 2009, 12:14, edited 3 times in total.
formulas
Math Expert
Joined: 02 Sep 2009
Posts: 29100
Followers: 4721

Kudos [?]: 49610 [3] , given: 7400

Re: GMATPrep DS Product of first 30 integers [#permalink]  30 Nov 2009, 21:54
3
KUDOS
Expert's post
If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?

(1) 10^d is a factor of f --> $$k*10^d=30!$$.

First we should find out how many zeros $$30!$$ has, it's called trailing zeros. It can be determined by the power of $$5$$ in the number $$30!$$ --> $$\frac{30}{5}+\frac{30}{25}=6+1=7$$ --> $$30!$$ has $$7$$ zeros.

$$k*10^d=n*10^7$$, (where $$n$$ is the product of other multiples of 30!) --> it tells us only that max possible value of $$d$$ is $$7$$. Not sufficient.

(2) $$d>6$$ Not Sufficient.

(1)+(2) $$d>6$$, $$d_{max}=7$$ --> $$d=7$$.

_________________
Manager
Joined: 14 Jul 2005
Posts: 104
Location: Sofia, Bulgaria
Followers: 1

Kudos [?]: 7 [2] , given: 0

2
KUDOS
Statement 1 tells us that we need to find out how many times is 30! divisible by 10. The hardest way to solve this is to break down 30! to its prime factors and count the 2s and 5s, because they make up the 10s. It is pretty easy to see that there are many more 2s than 5s in 30!, because we have 15 even numbers and only 6 numbers divisible by 5.

The numbers that contain 5s are 5=5, 2*5=10, 3*5=15, 4*5=20, 5*5=25, 6*5=30. So we have a total of seven 5s and more than seven 2s, which means that 30! can be evenly divided by 10 up to seven times. Therefore 1 <= d <=7. We can't figure out the exact value, so the statement is insufficient.

Statement 2 tells us that d > 6, which is a worthless piece of information on its own.

When we combine the 2 statements, we get C.

There was a very nice discussion of a similar problem about a month ago, but I can't find the post. The approach is "stolen" from there.
Math Expert
Joined: 02 Sep 2009
Posts: 29100
Followers: 4721

Kudos [?]: 49610 [2] , given: 7400

Re: Number Properties from GMATPrep [#permalink]  05 Oct 2009, 04:02
2
KUDOS
Expert's post
If you are aiming for 700+ in GMAT you should know 2 important things about factorials:

1. Trailing zeros:
Trailing zeros are a sequence of 0s in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

$$\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}$$, where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
$$\frac{32}{5}+\frac{32}{5^2}=6+1=7$$ (denominator must be less than 32, $$5^2=25$$ is less)

So there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

2. Finding the number of powers of a prime number k, in the n!.

What is the power of 3 in 35!

Tell me if you need this one too.
_________________
Director
Joined: 16 May 2007
Posts: 549
Followers: 2

Kudos [?]: 27 [1] , given: 0

DS : product of first 30 positive integers [#permalink]  09 Aug 2007, 18:39
1
KUDOS
If d is a positive integer and f is the product if the first 30 positive integers, what is the value of d?

1. 10^d is a factor of f.
2. d>6
VP
Joined: 10 Jun 2007
Posts: 1463
Followers: 6

Kudos [?]: 153 [1] , given: 0

Re: DS : product of first 30 positive integers [#permalink]  09 Aug 2007, 20:26
1
KUDOS
trahul4 wrote:
If d is a positive integer and f is the product if the first 30 positive integers, what is the value of d?

1. 10^d is a factor of f.
2. d>6

C.

Given: f=30!
(1) 10^d is a factor of f
Plug in d=1, 10 is a factor of f, Yes!
d=2, 100 is a factor of f, Yes! because 25*4 = 100
INSUFFICIENT.

(2) d>6. INSUFFICIENT

Together, plug in d=7, Is 10^7 is a factor of f?
5*2 = 10
10 = 10
15*12 = 180 = 18*10
20 = 2*10
25*4= 10*10
30 = 3*10
I don't think there is any more, SUFFICIENT.
Manager
Joined: 10 Jul 2009
Posts: 169
Followers: 1

Kudos [?]: 45 [1] , given: 8

Re: properties of intigers [#permalink]  10 Aug 2009, 10:21
1
KUDOS
f = 30!

1) 10^d is a factor of f
so we have to find the powers of 10 in the 30!
number of powers of 10 is equal to the number of 2 and 5
multiples of 5 less than or equal to 30 are 5,10, 15, 20, 25, 30.
So number of powers of 5 in 30! = 7
As we have many multiple is 2, the maximum value of d is 7
(i.e. d can be 1 or 2 or 3 or 4 ...)
2) d>6.
d can take any value.
Clubbing 1 and 2 we get,
d = 7
Manager
Joined: 08 Oct 2009
Posts: 66
Followers: 1

Kudos [?]: 21 [1] , given: 5

Re: Number Properties from GMATPrep [#permalink]  19 Oct 2009, 14:51
1
KUDOS
Great stuff Bunuel !!
Manager
Joined: 24 Jun 2009
Posts: 60
Followers: 1

Kudos [?]: 32 [1] , given: 2

GMATPrep DS Product of first 30 integers [#permalink]  30 Nov 2009, 17:44
1
KUDOS
Senior Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 312
Followers: 4

Kudos [?]: 120 [1] , given: 37

Re: GMATPrep DS Product of first 30 integers [#permalink]  30 Nov 2009, 19:07
1
KUDOS
I think the answer is C.

S1 by itself is not sufficient, coz if d=1 means 10 is a factor of 30!, true, if d =2, 100 is also a factor of 30!, d can be 1,2 or more... so insuff
S2 by itself is not sufficient, coz d>6 means d can be 7,8,9 or anything - clearly insuff

combining the two however we can asnwer the question, because in 30! we have 7 powers of 10 as below:

1.2.3.4.5 has one power for 10 (2*5)
6.7.8.9.10 has one power for 10 (10)
11.12.13.14.15 has one power for 10 (15*14 or 15*12)
16.17.18.19.20 has one power for 10 (20)
21.22.23.24.25 has 2 powers for 10 (25*24)
26.27.28.29.30 has one power for 10 (30)
total of 7 so $$10^7$$ is the highest $$10^d$$ being fact of 30! hence d=7
_________________

Thanks, Sri
-------------------------------
keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Math Expert
Joined: 02 Sep 2009
Posts: 29100
Followers: 4721

Kudos [?]: 49610 [1] , given: 7400

Re: DS : product of first 30 positive integers [#permalink]  11 Aug 2010, 15:26
1
KUDOS
Expert's post
masland wrote:
Is there anyway to quickly determine if d>7 is not a factor of 30! ?

If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?

(1) 10^d is a factor of f --> $$k*10^d=30!$$.

First we should find out how many zeros $$30!$$ has, it's called trailing zeros. It can be determined by the power of $$5$$ in the number $$30!$$ --> $$\frac{30}{5}+\frac{30}{25}=6+1=7$$ --> $$30!$$ has $$7$$ zeros.

$$k*10^d=n*10^7$$, (where $$n$$ is the product of other multiples of 30!) --> it tells us only that max possible value of $$d$$ is $$7$$. Not sufficient.

Side notes: 30! is some huge number with 7 trailing zeros (ending with 7 zeros). Statement (1) says that $$10^d$$ is factor of this number, but $$10^d$$ can be 10 (d=1) or 100 (d=2) ... or 10,000,000 (d=7). Basically $$d$$ can be any integer from 1 to 7, inclusive (if $$d>7$$ then $$10^d$$ won't be a factor of 30! as 30! has only 7 zeros in the end). So we can not determine single numerical value of $$d$$ from this statement. Hence this statement is not sufficient.

(2) $$d>6$$ Not Sufficient.

(1)+(2) $$d>6$$, $$d_{max}=7$$ --> $$d=7$$.

Hope it helps.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 29100
Followers: 4721

Kudos [?]: 49610 [1] , given: 7400

Re: DS : product of first 30 positive integers [#permalink]  13 Aug 2010, 01:53
1
KUDOS
Expert's post
estreet wrote:
It can be determined by the power of $$5$$ in the number $$30!$$ --> $$\frac{30}{5}+\frac{30}{25}=6+1=7$$ --> $$30!$$ has $$7$$ zeros.

I don't understand the calculations that were performed here. How did you get to $$\frac{30}{5}+\frac{30}{25}=6+1=7$$? How did you know that the 5 was the factor needed? Thanks

_________________
Intern
Joined: 09 Sep 2005
Posts: 21
Followers: 0

Kudos [?]: 2 [0], given: 0

WELL DONE ... BRILLIANT ... BRAVO

Many thanks
Manager
Joined: 06 Aug 2005
Posts: 197
Followers: 3

Kudos [?]: 6 [0], given: 0

The maximum value of d = int(d/5) + int (d/(5^2)) + ....

d = int(30/5) + int(30/25) + ... = 6 + 1 + 0 = 7
VP
Joined: 13 Jun 2004
Posts: 1119
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 33 [0], given: 0

this problem is great, I was totally lost and I would have gone for E
very nice answer Vasild, I am gonna study this later
Intern
Joined: 11 Sep 2005
Posts: 14
Followers: 0

Kudos [?]: 0 [0], given: 0

vasild wrote:
Statement 1 tells us that we need to find out how many times is 30! divisible by 10. The hardest way to solve this is to break down 30! to its prime factors and count the 2s and 5s, because they make up the 10s. It is pretty easy to see that there are many more 2s than 5s in 30!, because we have 15 even numbers and only 6 numbers divisible by 5.

The numbers that contain 5s are 5=5, 2*5=10, 3*5=15, 4*5=20, 5*5=25, 6*5=30. So we have a total of seven 5s and more than seven 2s, which means that 30! can be evenly divided by 10 up to seven times. Therefore 1 <= d <=7. We can't figure out the exact value, so the statement is insufficient.

Statement 2 tells us that d > 6, which is a worthless piece of information on its own.

When we combine the 2 statements, we get C.

There was a very nice discussion of a similar problem about a month ago, but I can't find the post. The approach is "stolen" from there.

i didnt get how we get "total of seven 5s"..i am able to see only six 5's.
Intern
Joined: 11 Sep 2005
Posts: 14
Followers: 0

Kudos [?]: 0 [0], given: 0

davesh wrote:
vasild wrote:
Statement 1 tells us that we need to find out how many times is 30! divisible by 10. The hardest way to solve this is to break down 30! to its prime factors and count the 2s and 5s, because they make up the 10s. It is pretty easy to see that there are many more 2s than 5s in 30!, because we have 15 even numbers and only 6 numbers divisible by 5.

The numbers that contain 5s are 5=5, 2*5=10, 3*5=15, 4*5=20, 5*5=25, 6*5=30. So we have a total of seven 5s and more than seven 2s, which means that 30! can be evenly divided by 10 up to seven times. Therefore 1 <= d <=7. We can't figure out the exact value, so the statement is insufficient.

Statement 2 tells us that d > 6, which is a worthless piece of information on its own.

When we combine the 2 statements, we get C.

There was a very nice discussion of a similar problem about a month ago, but I can't find the post. The approach is "stolen" from there.

i didnt get how we get "total of seven 5s"..i am able to see only six 5's.

My apologies..i got ur funda..
Manager
Joined: 26 Jun 2006
Posts: 152
Followers: 1

Kudos [?]: 1 [0], given: 0

GMATPrep DS: 10^d [#permalink]  19 Aug 2006, 09:15
Stumbled on this one. Anyone knows how to do it?

If d is a positive integer and F is the product of the first 30 positive integers, what is the value of d?

(1) 10^d is a factor of F

(2) d>6

Thanks!
Intern
Joined: 26 Apr 2005
Posts: 19
Followers: 0

Kudos [?]: 0 [0], given: 0

I think it is C
F = 1*2*3...*30
From A we know 10^d * X = F means..
and F contains for 7 instances of (5*2)
as in... 1*2*3*4...10 has two (5*2 and 10)
11 to 20 has (15 and 20)(meaning another 5 and 2*10 makes 2)
21 to 30 has a 25 and 30 (5 * 5 = 25 and 3 *10...for makes 3 instances of 10)
so d could be from 1 to 7...
from statement 2 u get that d > 6

Thus combining both u get the exact vlue of d...
Hence C

Go to page    1   2   3   4    Next  [ 70 posts ]

Similar topics Replies Last post
Similar
Topics:
2 If d is a positive integer, is d^1/2 an integer? 4 26 Feb 2014, 01:26
Is the product of a positive and a negative integer less 1 24 Aug 2012, 00:14
If c and d are positive, is d an integer ? 2 22 Jul 2012, 17:17
46 If d is a positive integer and f is the product of the first 17 28 Jan 2012, 17:13
4 If d is a positive integer, is d^1/2 an integer ? 6 15 Feb 2011, 10:53
Display posts from previous: Sort by