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24 Mar 2022, 10:50
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
70% (01:26) correct
30%
(01:49)
wrong
based on 542
sessions
History
Date
Time
Result
Not Attempted Yet
If \(N = (30!)^2\), what is the greatest integer \(k\) for which \(3^k\) is a divisor of \(N\)?
A) 14
B) 28
C) 42
D) 56
E) 196
A) 14
B) 28
C) 42
D) 56
E) 196
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29 Mar 2022, 13:55
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Basically we will need how many times 3 is present in (30!)^2. For this you can follow the method in factorials by bunuel. Basically it goes something like this
n/3 + n/3^2 + n/3^3 .... till 3^x is less than 30 . thus, it is
30/3 + 30/3^2 + 30/3^3 = 10+ 3+1 = 14
So this means we have 14 3's in 30!. But the questions asks for 30! square . That means there must another 14 in it. So its 14*2 =28
n/3 + n/3^2 + n/3^3 .... till 3^x is less than 30 . thus, it is
30/3 + 30/3^2 + 30/3^3 = 10+ 3+1 = 14
So this means we have 14 3's in 30!. But the questions asks for 30! square . That means there must another 14 in it. So its 14*2 =28
24 Mar 2022, 12:04
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Explanation:
(30!)^2
30/3 = 10
10/3 = 3
3/3 = 1
Sum = 10 + 3 + 1 = 14
As 30! is 2 times, Sum = 14x2 = 28
IMO-B
(30!)^2
30/3 = 10
10/3 = 3
3/3 = 1
Sum = 10 + 3 + 1 = 14
As 30! is 2 times, Sum = 14x2 = 28
IMO-B
