Totally confused!

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Find the number of trailing zeros in the expansion of (20!*21!*22! ......... *33!)^3!.

a) 10^468
b) 10^469
c) 10^470
d) 10^467
e) 10^471

Ans-A

[theCodeToGMAT]

Since the power is 3! = 6

So, the power of 10 must be divisible by 6

Only [spoiler]{A}[/spoiler]

[[email protected]]

Hi Rahul,


Dint get the logic!Please explain why

Thanks

Since the power is 3! = 6

So, the power of 10 must be divisible by 6

Only [spoiler]{A}[/spoiler]

[theCodeToGMAT]

Hi Rahul,


Dint get the logic!Please explain why

Thanks

See, in this question you can keep on solving question by solving each term and finding "0"s and then take power to 6.

GMAC doesn't want us to do so much calculation..that means there's some hidden trick

Analyze the answer choices.. you will see that they are in sequence.

For instance, if we put (10)^2 to the power of "6" so the term becomes 10^12 .. that means we are multiplying the power by another power... also, when the resultant power "12" is divided by the power to which we had raised the original term i.e. "6" we get remainder "0".

Similarly, coming back to original question, our final answer must have an answer choice which must have power divisible by "6".. Only answer choice {A} satisfies this.

I hope it's clear now.

[GMATGuruNY]

I received a PM asking me to comment.
The answer choices should read as I have posted them here:

Find the number of trailing zeros in the expansion of (20!*21!*22! ......... *33!)^3!.

a) 468
b) 469
c) 470
d) 467
e) 471

Let x = 20!*21! * ... *31!*33!.

Then:
(20!*21! * ... *31!*33!)^3! = x� = x * x * x * x * x * x.

If x*x*x*x*x*x is represented as an integer, the number of trailing 0's -- in other words, the number of 0's at the END of the integer -- will be equal to THE NUMBER OF 10'S contained within x*x*x*x*x*x.
Let n = the number of 10's contained within EACH x.
Then:
The 1st x contains n 10's.
The 2nd x contains n 10's.
The 3rd x contains n 10's.
The 4th x contains n 10's.
The 5th x contains n 10's.
The 6th x contains n 10's.
Total number of 10's = n+n+n+n+n+n = 6n.

Implication:
The total number of 10's -- and thus the total number of trailing 0's -- must be a MULTIPLE OF 6.
Of the answer choices, only A is a multiple of 6:
468/6 = 78.

The correct answer is A.

[theCodeToGMAT]

The more algebraic approach is:

Solve each term to get power of "10"

So,

20/5 => 4
21/5 => 4
22/5 => 4
23/5 => 4
24/5 => 4
25/5 => 5+1
26/5 => 5+1
27/5 => 5+1
28/5 => 5+1
29/5 => 5+1
30/5 => 6+1
31/5 => 6+1
32/5 => 6+1
33/5 => 6+1
So,
(4*5 + 6*5 + 7*4)^3! = So, (10^78)^6 = 10^468

[Matt@VeritasPrep]

A complete explanation is probably in order here:

Any time (2 * 5) appears in the prime factorization of a number, that number has a "trailing zero".

For example, consider the number 30. 30 = 3 * 2 * 5 = 3 * (2 * 5) = 3 * 10 = 30.

Now consider 300. 300 = 3 * 2 * 5 * 2 * 5 = 3 * (2 * 5) * (2 * 5) = 3 * 10 * 10 = 300.

Now consider 10!

10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
but we'll write it as
9 * 8 * 7 * 6 * 4 * 3 * 1 * 10 * 2 * 5

So 10! has two trailing zeros (! )

As you've probably gathered, any factorial is going to have more factors of 2 than it does of 5, so for every 5 we find we can easily find a 2 to pair with it.

Hence the question is really asking "How many 5's are there in the prime factorization of (20!*21!*22!*...*32!*33!)�?"

Now notice another shortcut. Every factorial here is a multiple of 20! (For instance, 21! = 21 * 20!.) So 20!, which has four factors of 15 (in its factors of 20, 15, 10, and 5, respectively) will have the same number of 5's as 21!, 22!, 23!, and 24!

So from 20! to 24!, we have 5 * 4 = 20 factors of 5.

25!, however, has two more 5's (since 25 = 5*5). 26! through 29! each have the same number of 5's in their factorizations, so we have another 5 * 6 = 30 factors of 5.

30! has another 5 in the 30, giving it seven factors of 5. 31!, 32!, and 33! are the same, so this gives us another 7 * 4 or 28 factors of 5.

So (20! * ... * 33!) has 20+30+28 = 78 factors of 5.

Since we're raising this to the 6th, we have 78*6 = 468 factors of 5, so we have 468 trailing zeros.