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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.
_________________

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.

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) \(d>6\), \(d_{max}=7\) --> \(d=7\).

Answer: C.

For trailing zeros see the link about factorials below.
_________________

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.
_________________

Re: DS : product of first 30 positive integers [#permalink]

Show Tags

09 Aug 2007, 21:26

1

This post received 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.

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 So answer is C

Re: GMATPrep DS Product of first 30 integers [#permalink]

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30 Nov 2009, 20:07

1

This post received 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

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) \(d>6\), \(d_{max}=7\) --> \(d=7\).

Answer: C.

For trailing zeros see the link about factorials in my signature.

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

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.

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.

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