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# 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at

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Math Expert
Joined: 02 Sep 2009
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30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

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12 Nov 2017, 00:48
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Question Stats:

38% (01:10) correct 62% (01:22) wrong based on 60 sessions

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$$30^{30} * 29^{29} * 28^{28}*...*2^2 * 1^1 = n$$. How many zeroes does ‘n’ contain at the end of the number (to the right of the last non-zero digit)?

A. 30
B. 60
C. 63
D. 105
E. 130
[Reveal] Spoiler: OA

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Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

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12 Nov 2017, 01:54
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The original expression can be rewritten as $$\frac{30!^{30}}{29!*28!*27!*...*3!*2!*1!}$$.
Using the forumla for factorials and the trailing zeroes:
$$\frac{30}{5} + \frac{30}{25} = 7$$
7 trailing zeroes for 30!, so the numerator has 210 zeroes.
Denominator has, using the same approach:
6 for 29!, 28!, 27!, 26! and 25!
4 for 24!, 23!, 22!, 21! and 20!
3 for 19!, 18!, 17!, 16! and 15!
2 for 14!, 13!, 12!, 11! and 10!
1 for 9!, 8!, 7!, 6! and 5!
0 for 4!, 3!, 2! and 1!.
Product of numbers with trailing zeroes makes these zeroes add up, then the resulting tally of zeroes in the denominator is: $$5*(6+4+3+2+1) = 5*16 = 80$$
Division with trailing zeroes makes you subtract the total of trailing zeroes in the denominator from the total of trailign zeroes in the numerator
$$210 - 80 =130$$

E

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Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

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12 Nov 2017, 02:22
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Count the number of 5's to get the number of Zeroes
Only 5,10,15,20,25,30 contains 5's
30^30 has 30 5's
25^25 has 50 5's
20^20 has 20 5's
15^15 has 15 5's
10^10 has 10 5's
5^5 has 5 5's
So the total is 130

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Math Expert
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Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

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12 Nov 2017, 03:33
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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)?

A. 30
B. 60
C. 63
D. 105
E. 130

Number of 10s will depend on number of 2s and 5s
Here since 5 is the larger prime number, it will give us the number of 0s
So number involving 5 are $$30^{30}*25^{25}*20^{20}*15^{15}*10^{10}*5^5=5^{30}*(5^2)^{25}*5^{20}*5^{15}*5^{10}*5^5=5^{30+2*25+20+15+10+5}=5^{130}$$
Ans 130
E
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

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12 Nov 2017, 04:32
To find the no. of zeros find either
1) no. of 5's in the series( as 5*2 =10 makes the zero)
2) if a no. contains the zeros itself (like 30, 20 and 10 in above series)

In the above series we have enough two's, all even no(28,26,24....2) will provide us the two. We have to only look for the 5's and that will come from 5, 15,25.
in 5^5 gives 5 5's
in 15^5 gives 15 5's
in 25^5 gives 50 5's

therefore total- 70 5's i.e. 70 zeros.

also, if a no. contain a zero itself when multiply itself it will have the no. of zeros equals to( no. of zeros in the no. at the right end)* (power of that no).

30^30 will give 30 zeros.
20^20 will give 20 zeros
10^10 will give 10 zeros

Total- 60 zeros

Overall we have then 60+70= 130 zeros.

Also see the attachment for more detail

Please hit the kudos button if you like this
Attachments

IMG_2.jpg [ 1.55 MiB | Viewed 565 times ]

IMG_1.jpg [ 1.72 MiB | Viewed 563 times ]

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Manager
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Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

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13 Nov 2017, 06:20
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)?

A. 30
B. 60
C. 63
D. 105
E. 130

Can someone explain this question in most easiest way???

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Intern
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Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at [#permalink]

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13 Nov 2017, 11:18
rocko911 wrote:
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)?

A. 30
B. 60
C. 63
D. 105
E. 130

Can someone explain this question in most easiest way???

Question is asking how many zeros will come at the right end if we mutiply the above series.

Suppose if we multiple : 101*100 then 10100 will be d answer and u can see before any digit other than zero appear we already have two zero at the right end .... So answer for this is two zeros....

Similarly for the above series we have to find how many continous zeros will come at right side before any other digit will appear ....

For d logic you can see the attached image ...

Just remember if we multiply any no. Than zero in the end will come if either the no. Has zero at the end in it like 10, 20, 300,. 400,....

Or if a the multiplying no. Has factor of 10 i.e one 5's and one 2's .... As 5*2 gives us 10

Eg 75 *8 ( in 75 we have 2 5's ( 3*5*5) and in 8 we have 3 tow's ) but we have only 2 fives and 3 twos so no. Of zero will come in the end is two)

75*8 = 600...

See above attachment and hit the kudos if u like this thanks

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Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at   [#permalink] 13 Nov 2017, 11:18
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# 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at

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