Alternate Solutions Needed

[knight247]

If the number 200! is written in the form p*10^q, where p and q are integers, what is the maximum possible value of q?
(A)40
(B)48
(C)49
(D)55
(E)64

OA is C

I'm kinda familiar with the standard way of solving along with the underlying principle of this problem i.e. finding the number of 5s in the entire factorial. To do this, we find the number of multiples of 5 which is 40, then number of multiples of 25 which is 8, and the number of multiples of 125 which is 1. Add them all together and u get 49.

Hoping to get an alternate solution to this one, if there is a better way to do it. Thanks

[Mike@Magoosh]

Hi, there. I'm happy to give my 2¢ on this.

In the BIG picture of all mathematical techniques, where calculators and computers are allowed, then yes, there would be other ways to approach the problems to solve for both p & q.

In the somewhat more restrictive picture of GMAT math, where you have no calculator, no technology, and just have to solve the problem with "pencil-and-paper" solutions, I'd say that the way you did this problem is the only pencil-and-paper" solution that occurs to me. We know 10 = 2*5, and the factors of 2 will be more plentiful than the factors of 5, so you have to count the factors of five. You did that flawlessly. The good news is: you know exactly how to solve the problem. The bad news is: unfortunately, there are no alternative GMAT-appropriate solutions to this problem. The further good news is: if you can do this, as you did, then you will be able to handle whatever other factorial problems the GMAT throws at you.

Here's another factorial question, for practice.

https://gmat.magoosh.com/questions/298

The question at that link should be followed by a full video solution.

I hope that was helpful. Please let me know if you have any further questions.

Mike

[pemdas]

If the number 200! is written in the form p*10^q, where p and q are integers, what is the maximum possible value of q?
(A)40
(B)48
(C)49
(D)55
(E)64

OA is C
I'm kinda familiar with the standard way of solving along with the underlying principle of this problem i.e. finding the number of 5s in the entire factorial. To do this, we find the number of multiples of 5 which is 40, then number of multiples of 25 which is 8, and the number of multiples of 125 which is 1. Add them all together and u get 49.

Hoping to get an alternate solution to this one, if there is a better way to do it. Thanks

To find the similar, trailing zeros, I will employ the exponential function y=200*5^n. Since in GMAT math sequence series the negative powers are not permitted and I am in need of the negative powers, I will be using calculus (note: this is beyond the scope of GMAT math) for alternative approach.

As such 200*5^-1 + 200*5^-2 ... finally 200*5^n when the exponential function y=200*5^n is close to zero, find function's limit -> 0, take derivative of the negative power of n and assign y -> 0. For the function y=200*5^n rewritten as y=8*5^(n+2), its derivative will be y`=(n+2)*5^(n+1). The function's limit->0 will be y`->0 and 5(n+1)*5^n->0 with the only possible n=-3, such as 5(-3+1)*5^-3=-10/125

Compare n=-4, 5(-4+1)*5^-4=-15/125 is farther from zero than -10/125

point n=-3 and assign trailing zeros as 200*5^-1 + 200*5^-2 + ... 200*5^n and n=-3

49 zeros total after any last non-zero digit.

[Johnkate]

The best way to solve such questions is to find the numbers which are factor of 5 and 5's power thereafter.

In 200!,

1. Number of direct factors of 5 = 200/5 = 40
2. Number of direct factors of 5*5, i.e. 25 = 200/25 = 8
3. Number of direct factors of 5*5*5, i.e. 125 = 200/125 = 1 (ignore the decimals)
4. Number of direct factors of 5*5*5*5, i.e 625 = 200/625 = 0

Once the result comes zero, do the total = 40 + 8 + 1 = 49

49 is the answer.

Enjoy!!!

[LalaB]

it is called to find trailing zeros . for more info please refer to the following link - https://en.wikipedia.org/wiki/Trailing_zero


200/5^1+200/5^2+200/5^3=40+8+1=49

we divide till 5^3, since 5^4 (625) is more than 200

[shankar.ashwin]

Well this isnt exactly an alternate method... But then its a fast way using the same concept..

Set up the number similar to finding factors of 5..

5 | 200
5 | 40 - 0
5 | 8 - 0
5 | 1 - 3

Leave out the remainders.. You have 40+8+1 = 49.