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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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Updated on: 26 May 2017, 06:29
2
4
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Difficulty:

55% (hard)

Question Stats:

63% (02:17) correct 37% (02:16) wrong based on 155 sessions

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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, 13:32.
Last edited by Bunuel on 26 May 2017, 06:29, edited 1 time in total.
Edited the question.
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Joined: 02 Aug 2009
Posts: 7099
Re: (30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = N  [#permalink]

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01 Feb 2017, 16:46
2
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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Re: (30^30) × (29^29) × (28^28) × . . . × (3^3) × (2^2) × (1^1) = N  [#permalink]

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02 Feb 2017, 01:16
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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12 Jun 2017, 19:28
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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09 Jul 2018, 05:19
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 &nbs [#permalink] 09 Jul 2018, 05:19
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