leading zeros

[gmatrant]

If P = (n)(n - 1)(n - 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


Shouldn't the answer be A.
The OA says C.
If you count the number of 5s in the number that should give you the number of leading zeros.
29 has six 5s. So thereby six leading zeros.

Can you please let me know if the above is correct. How is OA C?

[sanju09]

If P = (n)(n - 1)(n - 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


Shouldn't the answer be A.
The OA says C.
If you count the number of 5s in the number that should give you the number of leading zeros.
29 has six 5s. So thereby six leading zeros.

Can you please let me know if the above is correct. How is OA C?

If P has zero as its last 6 digits and a non-zero digit for its millions place, then 29 is not the smallest n that ensures it. It's 30 in fact.

2 zeros when 1 to 10 is multiplied out

2 more zeros when 11 to 20 is multiplied out

2 more zeros when 21 to 30 is multiplied out

Hence, 30 is the minimum n, a seventh zero would appear when n takes 35, till then there would remain six zeros as its last 6 digits when the largest n is [spoiler]34.


C[/spoiler]

[gmatrant]

The approach I tried was to count for number of 5s in the number.
Since a combination of 5 & 2 will give a 10, so thereby we can find leading zeros.

Take 30!
30! has seven 5s and more than seven 2s.
So we have seven 10s -> seven leading zeros.

Can you tell me what is wrong with this approach.

[gmatrant]

Math experts , can you please validate my response.

[spacealtgrctrl]

the answer is c.

30..25...22...20...15...12...10...5...2 when these are multiplied you get 6 zeros . so as per the options the 34 will give you the answer with a non zero in the million place...
pls correct me if i am wrong..

[sanju09]

The approach I tried was to count for number of 5s in the number.
Since a combination of 5 & 2 will give a 10, so thereby we can find leading zeros.

Take 30!
30! has seven 5s and more than seven 2s.
So we have seven 10s -> seven leading zeros.

Can you tell me what is wrong with this approach.

I rechecked, and found you as correct. [spoiler]A[/spoiler] should be the right answer here.

[sanju09]

the answer is c.

30..25...22...20...15...12...10...5...2 when these are multiplied you get 6 zeros . so as per the options the 34 will give you the answer with a non zero in the million place...
pls correct me if i am wrong..

gmatrant's approach and claim, both are correct; 25 contains 2 fives

[lokesh r]

the answer must be C.

We must count number of 5's.

3x4x5x..10x..x15..x20..x25..29

[puneetdua]

IMO - ans shd be A

1*10 -> 10 (1 zero)
2*5 -> 10 (1 zero)
15*6 -> 90 (1 zero)
20 --> (1 zero)
25 * 4 -> 100 (2 zero)

total of 6 zeros , if we increase our count to 30 - that will result in at least 7 zeros ...

So it shd be A.

What is the source of problem.
And if its official problem - how can we ask Expert to post their suggestions.

[lokesh r]

sorry..typing mistake..answer must be option A.29

[gmatrant]

Forum Math Experts - Can you please comment??

[Stuart@KaplanGMAT]

the answer must be C.

We must count number of 5's.

3x4x5x..10x..x15..x20..x25..29

As others have noted, 25 has two 5s, not just one.

So, when we calculate 25! (the definition for p is simply p=n!), we get 6 5s (5, 10, 15, 20, 25, 25) giving us a number that ends in 6 0s (since we have lots and lots of 2s).

We want the biggest number that gives us exactly 6 0s; since 30 will give us a 7th, the largest possible value for n is 29.

As always, please provide the source for all of your questions - where is this one from?

[gmatrant]

I don' recollect exactly.
I guess that this was given to me by a friend saying this was from a Princeton Math Quiz paper.

Thanks Stuart and all for confirming the answer.