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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # 30^30 x 29^29 x 28^28…2^2 x 1^1 = n.  Question banks Downloads My Bookmarks Reviews Important topics
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TAGS: Senior Manager Joined: 20 May 2014
Posts: 282
Followers: 18

Kudos [?]: 50  , given: 220

30^30 x 29^29 x 28^28…2^2 x 1^1 = n. [#permalink]
1
KUDOS 00:00

Question Stats: 100% (03:05) correct 0% (00:00) wrong based on 3 sessions
$$30^{30} * 29^{29} * 28^{28}*...*2^2 * 1^1 = n$$. How many zeroes does ‘n’ contain at the end of the number (to the right of the last non-zero digit)?

[Reveal] Spoiler: OA
130
Intern Joined: 14 Oct 2017
Posts: 2
Followers: 0

Kudos [?]: 0 , given: 1

Re: 30^30 x 29^29 x 28^28…2^2 x 1^1 = n. [#permalink]
Hi
Taking the maximum power of 2 and 5
2^(2+4+6+8+.......30) and 5^(5+10+15+ ....+30)
power of 2 is 240 & power of 5 is 105
number of zeros to the end of the number is 105.
Can you please let me know where I'm doing wrong
Senior Manager Joined: 20 May 2014
Posts: 282
Followers: 18

Kudos [?]: 50 , given: 220

Re: 30^30 x 29^29 x 28^28…2^2 x 1^1 = n. [#permalink]
Bunuel wrote:
$$30^{30} * 29^{29} * 28^{28}*...*2^2 * 1^1 = n$$. How many zeroes does ‘n’ contain at the end of the number (to the right of the last non-zero digit)?

[Reveal] Spoiler: OA
130

Magoosh Answer and Explanation

Some of the zeroes are easy to get: the one’s in which the number is a multiple of 10. So 30^30 will yield 30 zeroes, 20^20 will yield 20 zeroes, and 10^10 will yield 10 zeroes, giving us a total of sixty zeroes.

But here is where things start to get trickier. Don’t forget the ‘5’s and the ‘2’s. Why? Well, as long as we can factor out a combination of a ‘5’ and a ‘2’, we have 5 x 2 = 10, meaning exactly one zero. Now before you go counting all the ‘2’ in this enormous number—which might take all day—it is better to get the number of ‘5’s’ since there are fewer and you know that however many ‘5’s you count there will be enough ‘2’s to pair it off with.

5^5 = five zeroes

15^15 = fifteen zeroes

25^25 = ….here’s the doozy. 25 = 5^2. Therefore, 25^25 = 5^50, giving us a whopping fifty zeroes.

So let’s add these up:

5 + 15 + 50 = 70

Adding these to the 60 zeroes we got from all those factors of 10 gives us 130, the answer. Re: 30^30 x 29^29 x 28^28…2^2 x 1^1 = n.   [#permalink] 26 Nov 2017, 08:57
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