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OA is ......
C
The question for 2 is d > 6, no typo
Last edited by lfox2 on 01 Jan 2007, 04:09, edited 1 time in total.
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Fig wrote: (C) for me  f = 30! To me, I decompose 10 in prime factor. 10 = 5*2. Obviously, in 30!, there is a lot of "2" prime factors. So, the limit comes from the number of 5 prime factors. 5 is contained in : 5, 10, 15, 20, 25, 30. So, we have 6 5 prime factors available. From 1f = k*10^d Following what we said about 5, d could be anything from 1 to 6. INSUFF. From 2d > 6 : d = 7 or d = 8. It gives nothing alone. INSUFF. Both (1) and (2)We know the maximum is 6... Is it a typo? I would say d >= 6. Because d cannot be 7. But we can conclude impossible SUFF. No typo :p... I have forgotten 25 = 5*5 ... 7 5 are available 
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c for me as well.
10 has 2 and 5 as factors,
2 occurs
30 / 2 = 15;
15/2 = 7;
7/2 = 3
3/2 = 1;
(15 + 7 + 3 + 1) = 26 times
While 5 occurs,
30/ 5 = 6
6/5 = 1
6 + 1 = 7 times
there for 10 will have 7 instances in 30!
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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.
(2) d>6
Can someone suggest a good technique for this one please!
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doc14 wrote: 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
Can someone suggest a good technique for this one please!
The best way to solve problems like that is to brake it into small pieces:
f = 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20*
21*22*23*24*25*26*27*28*29*30
First off - try to convert the number above into something that you can work with - look at the statements first to get more info.
I left only the numbers that are multiplication of 10 !
2*4*5*10*20*15*25*30 = convert into = 2*2*5*10*2*10*3*5*5*5*10*3
get rid off the 3 = 10*10*10*10*10*10*10 (total of 7)
now the magic begins !
statement 1
we were asked about 10^n as a factor of - f - so look up - there no way to know - d can be 1,2,3,4,5,6,7
insufficient
statement 2
dosen't say much by itself
insufficient
statment 2&1
since there is 7 exponents of 10 that are a factor of - f
if n>6 n has to be 7.
sufficient
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I think the best way to solve the problem is this:
10 can only be made up of 2s and 5s (2x5). In the list of 1x2x3...x30, there are plenty of 2s, but only 7 5s (5, 10, 15, 20, 30 have one each, and 25 has 2 5s). So, f has 7 10s .. ie, f = k x 10^7
So, from statement one, d can be anything from 1 to 7
From statement two, d should be >6
Neither of the statements are sufficient in themselves, but together, they make d = 7
Hence C
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f = 30!
St1:
10^d is a factor of f. d can be 1,2,3 etc... Insufficient.
St2:
Uesless. We know nothing much else. Insufficient.
St1 and St2:
If d = 7, then we have 10^7 = (2^7)*(5^7). That's all the 5's we have in 30!. So d cannot be 8 and up. Sufficient.
Ans C
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OA is C.
Thanks for all the replies!
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DS : product of first 30 positive integers [#permalink]
 09 Aug 2007, 19:39
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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
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Re: DS : product of first 30 positive integers [#permalink]
09 Aug 2007, 21:26
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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.
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Re: DS : product of first 30 positive integers [#permalink]
10 Aug 2007, 01:23
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.
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properties of intigers [#permalink]
10 Aug 2009, 11:03
If d is a positive integer and f is the prodcut of the first 30 positive integers, what is the value of d?
(1) 10^d is a factor of f
(2) d > 6
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Re: properties of intigers [#permalink]
10 Aug 2009, 11:21
have read it before and found the soln here ... - http://www.manhattangmat.com/forums/if- ... t2234.htmltough one ... i will get screwed if something like this appears :D
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Re: properties of intigers [#permalink]
10 Aug 2009, 11:21
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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
So answer is C
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Number Properties from GMATPrep [#permalink]
04 Oct 2009, 01:54
Please help me to solve the problem.
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Re: Number Properties from GMATPrep [#permalink]
04 Oct 2009, 06:26
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Re: Number Properties from GMATPrep [#permalink]
04 Oct 2009, 20:56
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 could you explain the trailing zeros bit in more detail please?
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Re: Number Properties from GMATPrep [#permalink]
04 Oct 2009, 22:06
yangsta8 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 could you explain the trailing zeros bit in more detail please? Please explain the trailing zeroes part!
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Re: Number Properties from GMATPrep [#permalink]
05 Oct 2009, 05:02
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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)>nIt'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.
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Re: Number Properties from GMATPrep [#permalink]
05 Oct 2009, 05:43
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?
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Re: Number Properties from GMATPrep
[#permalink]
05 Oct 2009, 05:43
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