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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
Posts: 60778
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, 01:48
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95% (hard)

Question Stats:

36% (02:18) correct 64% (01:58) wrong based on 130 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

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Posts: 8322
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, 04:33
4
5
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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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, 02:54
1
4
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, 03:22
3
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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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, 05:32
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
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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, 07:20
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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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, 12:18
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]

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29 Dec 2017, 08:11
Basically we have to find number of 5s in n

30^30 : 30 is multiple of 5 => so we have 30 5s
25^25 : 25 is mulitple of 5 => so we have 25 5s
20^20 will have =>20 5s
15^15 will have =>15 5s
10^10 will have =>10 5s
5^5 will have => 5 5s
total = 105 5s

but we are not finished yet, 25 has two 5s in it
so we have to do above iteration, to check how many 25s are there
only 25^25 has 25 => 25^25 will have 25 5s (note: we are counting only once, because another 5 in 25 is counted in above iteration)
total = 25 5s

we could also perform the above iteration for 5^3 meaning how many 125 are there, but we have only till 30, so we can stop here.

n has altogether, 130 5s => ofcourse, no of 2s will be greater than number of 5s

but forming a 10 is restricted by number of 5s, so n has 130 10s, => n will have 130 zeros at the end of rightmost non-zero digit => Answer (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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08 Dec 2019, 04:35
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Re: 30^30*29^29*28^28*...*2^2*1^1 = n. How many zeroes does ‘n’ contain at   [#permalink] 08 Dec 2019, 04:35
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