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(30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = N

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(30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = N [#permalink]

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New post Updated on: 26 May 2017, 07:29
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\((30^{30}) × (29^{29}) × (28^{28}) × . . . × (3^3) × (2^2) × (1^1) = N\)

What is the highest value of K, such that \(\frac{N}{(125^K)}\) is an integer?

A) 27
B) 35
C) 38
D) 43
E) 48

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Originally posted by daboo343 on 01 Feb 2017, 14:32.
Last edited by Bunuel on 26 May 2017, 07:29, edited 1 time in total.
Edited the question.
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Re: (30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = N [#permalink]

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New post 01 Feb 2017, 17:46
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daboo343 wrote:
(30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = N

What is the highest value of K, such that N/(125^K) is an integer?

A) 27
B) 35
C) 38
D) 43
E) 48



Hi,
125 =5^3, so we have to find number of 5s in \((30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = N\)..
Multiples of 5 are \(5^5*10^{10}*15^{15}*20^{20}*25^{25}*30^{30}........5^5*5^{10}*5^{15}*5^{20}*(5^2)^{25}*5^{30}=5^{5+10+15+20+50+30}=5^{130}\)
But we are looking for powers of 125 or 5^3... So 130/3=43 1/3..
Thus ans is 43
D
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: (30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = N [#permalink]

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New post 02 Feb 2017, 02:16
daboo343 wrote:
(30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = N

What is the highest value of K, such that N/(125^K) is an integer?

A) 27
B) 35
C) 38
D) 43
E) 48



D..
total 5 in the series -5^(5+10+15+20+50+30)=5^130.
Denominator we can 125^k=(5*5*5)..43 will give 129.
So D .
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Re: (30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = N [#permalink]

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New post 12 Jun 2017, 20:28
Worked out the sum of powers as 130, but divided by 2 instead of 3!
Great explanations!
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Re: (30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = N [#permalink]

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New post 09 Jul 2018, 06:19
Quote:
\(N = 30^{30} * 29^{29} * 28^{28} * . . . * 3^3 * 2^2 * 1^1\)

What is the greatest possible integer value for \(k\) such that \(\frac{N}{125^k}\) is an integer?

A) 27
B) 35
C) 38
D) 43
E) 48


\(125^k = 5^{3k}\)

\(N = 30^{30} * 29^{29} * 28^{28} * . . . * 3^3 * 2^2 * 1^1\)
To determine how many times 5^{3k} can divide into N, count how many 5's are contained within the product directly above.

\(30^{30} = 5^{30}6^{30}\) --> thirty 5's.
\(25^{25} = (5^2)^{25} = 5^{50}\) --> fifty 5's.
\(20^{20} =5^{20}4^{20}\) --> twenty 5's.
\(15^{15} = 5^{15}3^{15}\)--> fifteen 5's.
\(10^{10} = 5^{10}2^{10}\) --> ten 5's.
\(5^5\) = five 5's.
Total number of 5's = \(30+50+20+15+10+5 = 130\).

Implication:
\(N = 5^{130}a\), where \(a\) = the product of the prime factors other than 5.
Since \(a\) is not divisible by 5 -- rendering the value of \(a\) irrelevant -- the question becomes:
If \(\frac{5^{130}}{5^{3k}}\) is an integer, what is the greatest possible integer value for \(k\)?

The division above will yield an integer if the exponent in the denominator is less than or equal to the exponent in the numerator:
\(3k < 130\)
\(k < 130/3\)
\(k < 43.33\)

Thus, the greatest possible integer value for \(k\) is 43.


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Re: (30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = N   [#permalink] 09 Jul 2018, 06:19
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