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29 Jan 2019, 03:55
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
57% (02:06) correct
43%
(02:03)
wrong
based on 227
sessions
History
Date
Time
Result
Not Attempted Yet
What is the highest positive integer power of 7 in 5000!?
(A) 714
(B) 816
(C) 832
(D) 835
(E) 4998
(A) 714
(B) 816
(C) 832
(D) 835
(E) 4998
What is the highest power of 7 in 5000!?
29 Jan 2019, 04:03
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Bunuel
CONCEPT: Power of any Prime Number in any factorial can be calculated by following understanding Power of prime x in\(n! = [n/x] + [n/x^2] + [n/x^3] + [n/x^4] +\)... and so on Where,
[\(n/x\)] = No. of Integers that are multiple of x from 1 to n
[\(n/x^2\)] = No. of Integers that are multiple of x^2 from 1 to n whose first power has been counted in previous step and second is being counted at this step
[\(n/x^3\)] = No. of Integers that are multiple of x^3 from 1 to n whose first two powers have been counted in previous two step and third power is counted at this step
And so on.....
Where [\(n/x\)] is greatest Integer value of (n/x) less than or equal to (n/x) i.e. [100/3] = [33.33] = 33 i.e. [100/9] = [11.11] = 11 etc.
i.e. Power of 7 in 5000! \(= [5000/7] + [5000/7^2] + [5000/7^3] + [5000/7^4] + ...\)
i.e. Power of 7 in 5000! \(= 714 + 102 + 14 + 2 = 832\)
Answer: Option C
Archit3110
29 Jan 2019, 04:14
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Bunuel
5000!/7 + 5000/7^2 + 5000/7^3 + 50007^4
= 714+102+14+2
832
IMO C
