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

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Re: Number Properties from GMATPrep [#permalink]

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05 Oct 2009, 04:58
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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.
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Last edited by bb on 21 Oct 2009, 12:14, edited 3 times in total.
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Re: Number Properties from GMATPrep [#permalink]

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17 Oct 2009, 14:14
Bunuel, you are the man!!! Absolutely briliant!! +1

Btw, where did you learn these formulas/tricks?
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Re: Number Properties from GMATPrep [#permalink]

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19 Oct 2009, 14:51
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Great stuff Bunuel !!
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Re: Number Properties from GMATPrep [#permalink]

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19 Oct 2009, 17:22
eresh wrote:
Bunuel, you are the man!!! Absolutely briliant!! +1

Btw, where did you learn these formulas/tricks?

Bunuel! the "Math Guru"! +1 for you...
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Re: Number Properties from GMATPrep [#permalink]

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20 Oct 2009, 09:12
Hi Mr Guru!!

Where does this kind of tips can be found ???

Honestly... I would have been stuck with such a problem. I would have answered probably C (first guess) but without knowing why.

Thanks for your feedback.
regards,
Alex
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Re: Number Properties from GMATPrep [#permalink]

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20 Oct 2009, 17:44
pierrealexandre77 wrote:
Hi Mr Guru!!

Where does this kind of tips can be found ???

Honestly... I would have been stuck with such a problem. I would have answered probably C (first guess) but without knowing why.

Thanks for your feedback.
regards,
Alex

Yes. Please share your knowledge.
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GMATPrep DS Product of first 30 integers [#permalink]

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30 Nov 2009, 17:44
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Re: GMATPrep DS Product of first 30 integers [#permalink]

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30 Nov 2009, 19:07
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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
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Re: GMATPrep DS Product of first 30 integers [#permalink]

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30 Nov 2009, 21:54
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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.
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Help! [#permalink]

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26 Jul 2010, 10:24
If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?
A. 10^d is a factor of f
B. d>6
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Re: GMATPrep DS Product of first 30 integers [#permalink]

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27 Jul 2010, 01:39
Answer is C. Yes we should use the formula to calculate the number of zeros, as does Bunuel. It is very helpful.
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Re: Number Properties from GMATPrep [#permalink]

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02 Aug 2010, 23:54
Bunuel wrote:
(1) k*10^d=30!
First we should find out how many zeros does 30! has, it's called trailing zeros. It can be determined by the power of 5 in the number 30! --> 1 in 5, 1 in 10, 1 in 15, 1 in 20, 2 in 25=(5*5) , 1 in 30 --> 1+1+1+1+2+1=7 --> 30! has 7 zeros.
k*10^d=n*10^7, n>=k it tells us only that max possible value of d is 7. Not Sufficient.

(2) d>6 Not Sufficient.

(1)+(2) d>6, dmax=7 --> d=7

C.

Bunuel, I did not get why k*10^d=n*10^7, n>=k is the case? why is n>=k? If we said k*10^d = n*10^7, then why is k=n and d=7 wrong?
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Re: Number Properties from GMATPrep [#permalink]

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03 Aug 2010, 13:11
mainhoon wrote:
Bunuel wrote:
(1) k*10^d=30!
First we should find out how many zeros does 30! has, it's called trailing zeros. It can be determined by the power of 5 in the number 30! --> 1 in 5, 1 in 10, 1 in 15, 1 in 20, 2 in 25=(5*5) , 1 in 30 --> 1+1+1+1+2+1=7 --> 30! has 7 zeros.
k*10^d=n*10^7, n>=k it tells us only that max possible value of d is 7. Not Sufficient.

(2) d>6 Not Sufficient.

(1)+(2) d>6, dmax=7 --> d=7

C.

Bunuel, I did not get why k*10^d=n*10^7, n>=k is the case? why is n>=k? If we said k*10^d = n*10^7, then why is k=n and d=7 wrong?

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.

Hope it's clear.
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Re: DS : product of first 30 positive integers [#permalink]

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11 Aug 2010, 15:17
GK_Gmat wrote:
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

I get C as well.

Stat1: d =1 or 2; insuff

Stat2: insuff

Stat 1&2: d=7, suff.

If 10^d is a factor of 30! and d>6 then d has to be 7 since 10^d is not a factor of 30! if d>7.

Is there anyway to quickly determine if d>7 is not a factor of 30! ?
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Re: DS : product of first 30 positive integers [#permalink]

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11 Aug 2010, 15:25
Hmm think I might have got it.

30! has only 7 5s in it...

5, 10, 15, 20, 25 (2 fives), 30.
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Re: DS : product of first 30 positive integers [#permalink]

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11 Aug 2010, 15:26
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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) $$d>6$$, $$d_{max}=7$$ --> $$d=7$$.

Answer: C.

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

Hope it helps.
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Re: DS : product of first 30 positive integers [#permalink]

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11 Aug 2010, 18:06
masland wrote:
Hmm think I might have got it.

30! has only 7 5s in it...

5, 10, 15, 20, 25 (2 fives), 30.

Yep, this is the easiest way to do it. 10^7 can't be a factor of 30! b/c 10^7 has more factors of 5 than 30!.

Whenever you see biiiig number like this and the word "factor" in the problem, this should be a big hint that prime factorization will get you to the final answer relatively quickly. Also, since 10^[any int] will only have prime factors of 2 and 5, and without much looking you can see that 30! will have a whole h*ll of a lot of factors of 2, you might be able to deduce quickly that the factor of 5 is going to be the limiting element here. So, figure out how many factors of 5 30! has in it, and you're made in the shade.
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Re: DS : product of first 30 positive integers [#permalink]

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11 Aug 2010, 18:07
madammepsychosis wrote:
masland wrote:
Hmm think I might have got it.

30! has only 7 5s in it...

5, 10, 15, 20, 25 (2 fives), 30.

Yep, this is the easiest way to do it. 10^7 can't be a factor of 30! b/c 10^7 has more factors of 5 than 30!.

Whenever you see biiiig number like this and the word "factor" in the problem, this should be a big hint that prime factorization will get you to the final answer relatively quickly. Also, since 10^[any int] will only have prime factors of 2 and 5, and without much looking you can see that 30! will have a whole h*ll of a lot of factors of 2, you might be able to deduce quickly that the factor of 5 is going to be the limiting element here. So, figure out how many factors of 5 30! has in it, and you're made in the shade.

Amend that: 10^7 does have enough 5's, but 10^8 wouldn't.
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Re: DS : product of first 30 positive integers [#permalink]

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12 Aug 2010, 17:46
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

Bunuel wrote:
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) $$d>6$$, $$d_{max}=7$$ --> $$d=7$$.

Answer: C.

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

Hope it helps.
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Re: DS : product of first 30 positive integers [#permalink]

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13 Aug 2010, 01:53
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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

For trailing zeros see the link about factorials: everything-about-factorials-on-the-gmat-85592.html
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Re: DS : product of first 30 positive integers   [#permalink] 13 Aug 2010, 01:53

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