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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