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

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

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10 Sep 2005, 12:58
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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 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

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10 Sep 2005, 15:13
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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.

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10 Sep 2005, 16:27
WELL DONE ... BRILLIANT ... BRAVO

Many thanks

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10 Sep 2005, 22:04
The maximum value of d = int(d/5) + int (d/(5^2)) + ....

d = int(30/5) + int(30/25) + ... = 6 + 1 + 0 = 7

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10 Sep 2005, 22:44
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

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15 Sep 2005, 14:18
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.

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15 Sep 2005, 14:20
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..

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19 Aug 2006, 10:15
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!

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19 Aug 2006, 10:35
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

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19 Aug 2006, 10:48
Got it! After carefully tracing back my steps I reached the same conclusion.

I actually approached this problem the same way you did, but I counted way too many 10s in 30! (Somehow 20 produced two 10s for me instead of one!)

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I CANT GET IT (INTIGERS AND FACTORIALS) [#permalink]

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04 Sep 2006, 11:53
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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04 Sep 2006, 12:17
This question is testing ur understanding of number properties...

f=30!

D is a factor of f..

(1) 10^d is a factor of F....

well..10 is a fator of F, so is 100...Insuff

think of it this way...what are the prime factors of 10?

10=2*5, then 10^d, will have prime factors=2^a * 5^b..

now..we know that there are going to be fewer 5s in 30 factorial...so whatever the power of 5 is...will determine the power of 10^d.

in other words 5^b=10^d (the biggest possible value of 10 tht is a factor of 30!)

(2) d is greater than 6...

well that by itself is insuff

together Sufficient...

here is how..

30/5=6 fives...
25 has 2 fives, so there is one additional 5...

so we know that 5^b; where b=7...

so ..10^7 is the highest possible factor of 10 in 30!...

Sufficient...

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04 Sep 2006, 12:32
Thanks a lot ... i appreciate the time you gave me

to write this detailed explanation

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26 Nov 2006, 13:02
If d is a positive integer and f is the product of the first 30 positive integers, what's the value of d?

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

I did this Q right, however when I checked the factorial of "f" in Excel I got a different result... is Bill gates wrong
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Last edited by ugo_castelo on 27 Nov 2006, 04:26, edited 1 time in total.

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27 Nov 2006, 01:23
ugo_castelo wrote:
If d is a positive integer and f is the product of the first 30 integers, what's the value of d?

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

I did this Q right, however when I checked the factorial of "f" in Excel I got a different result... is Bill gates wrong

What does first 30 integers mean? What is the FIRST Integer anyway?

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27 Nov 2006, 04:26
yes, it's Positive... integers ( typed too fast)
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27 Nov 2006, 09:58
If d is a positive integer and f is the product of the first 30 positive integers, what's the value of d?

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

1st 30 = 1*2*3*4*... *30
we know this has 5*10*15*20*30 so 10^1 is fcator 10^2 is factor
(1) 10^d is factor insuff
(2) insuff on its own

Togather

Lets check how many multiple 5s are in product

5*10*15*20*25*30
5*5*2*5*3*5*4*5*5*5*6 so 5^7
we also have many 2s 2,4,8,... surely we can find 7 or more

5^7*2*7=10^7

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27 Nov 2006, 14:30
that's correct
the same way I did
However , when I checked the Factorial of 30 in excel, I got more then 10^7

that's why the surprise
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01 Jan 2007, 02:36
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

What is the best way of handling this type of question

OA to follow

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01 Jan 2007, 03:57
(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 1
f = k*10^d
Following what we said about 5, d could be anything from 1 to 6.

INSUFF.

From 2
d > 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.

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01 Jan 2007, 03:57

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