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02 Dec 2016, 06:58
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
80% (00:31) correct
20%
(00:43)
wrong
based on 167
sessions
History
Date
Time
Result
Not Attempted Yet
What is the greatest value of integer n such that 5^n is a factor of 15! ?
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
02 Dec 2016, 08:07
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From 15! values which will have value 5 in numbers are 15, 10 and 5.
15 will have one 5, 10 will have one 5 and 5 itself.
Hence, greatest value of integer n such that 5^n is a factor of 15! = 3. I hope that is the answer.
15 will have one 5, 10 will have one 5 and 5 itself.
Hence, greatest value of integer n such that 5^n is a factor of 15! = 3. I hope that is the answer.
22 Nov 2019, 17:06
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Bunuel
-----ASIDE---------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If k is a factor of N, then k is "hiding" within the prime factorization of N
Consider these examples:
3 is a factor of 24, because 24 = (2)(2)(2)(3), and we can clearly see the 3 hiding in the prime factorization.
Likewise, 5 is a factor of 70 because 70 = (2)(5)(7)
And 8 is a factor of 112 because 112 = (2)(2)(2)(2)(7)
And 15 is a factor of 630 because 630 = (2)(3)(3)(5)(7)
-----BACK TO THE QUESTION!---------------------
What is the greatest value of integer n such that 5^n is a factor of 15! ?
In other words, "How many 5's are hiding within the prime factorization of 15!"
15! = (15)(14)(13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)
= (3)(5)(14)(13)(12)(11)(2)(5)(9)(8)(7)(6)(5)(4)(3)(2)(1)
There are three 5's in the prime factorization of 15!
In other words, 5³ is a factor of 15!
Answer: C
Cheers,
Brent
