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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
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
01 Feb 2017, 17:46
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daboo343
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 \(\frac{130}{3}=43 \frac{1}{3}\).
Thus, the answer is 43
D
Updated on: 06 Nov 2020, 12:07
Originally posted by GMATGuruNY on 09 Jul 2018, 06:19.
Last edited by GMATGuruNY on 06 Nov 2020, 12:07, edited 1 time in total.
Last edited by GMATGuruNY on 06 Nov 2020, 12:07, edited 1 time in total.
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Quote:
\(125^k = 5^{3k}\)
\(N = 30^{30} * 29^{29} * 28^{28} * . . . * 3^3 * 2^2 * 1^1\)
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.
