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# If P = (n)(n 1)(n 2) . . . (1) and n > 2, what is the

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Manager
Joined: 18 Jun 2004
Posts: 105
Location: san jose , CA
Followers: 1

Kudos [?]: 32 [0], given: 0

If P = (n)(n 1)(n 2) . . . (1) and n > 2, what is the [#permalink]  31 Aug 2004, 01:51
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If P = (n)(n â€“ 1)(n â€“ 2) . . . (1) and n > 2, what
is the largest value of integer n where P has zero
as its last 6 digits and a non-zero digit for its
millions place?
(A) 29
(B) 30
(C) 34
(D) 35
(E) 39
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Manager
Joined: 02 Apr 2004
Posts: 224
Location: Utrecht
Followers: 1

Kudos [?]: 13 [0], given: 0

I will be very honest and I do not have any idea how to solve this problem. Completely blank!!!!

I would make an guess on the real test and this would be 30. The reason is that maybe because the last digit is zero I will get 6 zero digits.

I hope someone has an clear explanation for this question.

Regards,

Alex
Senior Manager
Joined: 19 May 2004
Posts: 291
Followers: 1

Kudos [?]: 9 [0], given: 0

B.

Alex, on first look i was thinking: Ha ??!

But then i realized that i should be looking for a factorial,
which contains at least 6 multiples of 5.
The numer 30 (6*5) is the least number with this property.
30! contains 6 multiples of 5: 5,10,15,20,25,30
And more than 6 multiples of 2.
Therefore this number should end with 6 zeros.

( i hope this is the correct answer , if it is.. good guess !)
CIO
Joined: 09 Mar 2003
Posts: 463
Followers: 2

Kudos [?]: 35 [0], given: 0

Dookie wrote:
B.

Alex, on first look i was thinking: Ha ??!

But then i realized that i should be looking for a factorial,
which contains at least 6 multiples of 5.
The numer 30 (6*5) is the least number with this property.
30! contains 6 multiples of 5: 5,10,15,20,25,30
And more than 6 multiples of 2.
Therefore this number should end with 6 zeros.

( i hope this is the correct answer , if it is.. good guess !)

Oh, no! Dookie, man, that was brilliant. I was lost on this at first, and I scrolled down, saw the first line of your response, and took off with it. You're totally right - it must have 6 fives and 6 twos.

BUT 25 had 2 5's! So 29 has it already! 30 takes it too far - with 30 you'll have 7 fives and all those two's - you're up by another factor of 10.

So the answer has to be A.

The confusing thing was that this problem insinuates that it's got exactly 7 digits, or at least, that's how I read it. But in fact, it's got plenty more than that...
SVP
Joined: 16 Oct 2003
Posts: 1813
Followers: 4

Kudos [?]: 56 [0], given: 0

Well I got 29 also. I may have done something non-sensical.

I started reverse from 1 ......n

1
2
3
4
5
6
7
8
9
10

This will have a max of 2 zeros 5*2*10 = 100
From 11 to 20 (12*15*20) again two zeros = .....00
From 21 to 25 again 2 zeros .....00.

so a total of six zeros. n=25 gives us what we need but can go till 29 since multiplying it by 30 means one more 0 with a total of 7 zeros. we just need 6 zeros.
Senior Manager
Joined: 19 May 2004
Posts: 291
Followers: 1

Kudos [?]: 9 [0], given: 0

oops, fell in the trap!

Thanks Ian.
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# If P = (n)(n 1)(n 2) . . . (1) and n > 2, what is the

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