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Updated on: 12 Dec 2018, 04:01
Originally posted by EgmatQuantExpert on 12 Dec 2018, 03:37.
Last edited by EgmatQuantExpert on 12 Dec 2018, 04:01, edited 4 times in total.
Last edited by EgmatQuantExpert on 12 Dec 2018, 04:01, edited 4 times in total.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
70% (02:17) correct
30%
(01:52)
wrong
based on 539
sessions
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Variations in Factorial Manipulation - Exercise Question #1
The value of the number F is equal to the product of all the numbers from 2 to 89. What is the value of the greatest integer n, so that \(12^n\) is a factor of F?
To solve question 2: Question 2
To read the article: Variations in Factorial Manipulation
The value of the number F is equal to the product of all the numbers from 2 to 89. What is the value of the greatest integer n, so that \(12^n\) is a factor of F?
- A. 5
B. 7
C. 12
D. 42
E. 85
To solve question 2: Question 2
To read the article: Variations in Factorial Manipulation
14 Dec 2018, 21:51
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Solution
Given:
- • F is a number whose value is equal to the product of all the numbers from 2 to 89.
- o Hence, F is equal to 89!
- o In other words, n is the highest power of 12 that can divide F.
To find:
- • The value of the greatest integer n, for which \(12^n\) is a factor of F.
Approach and Working:
As 12 is a composite number, first we need to prime factorize it in terms of the prime factors and their corresponding powers.
- • \(12 = 2^2 * 3^1\)
Next, we need to find the individual instances of 2 and 3.
- • Instances of 2: \(\frac{89}{2^1} + \frac{89}{2^2} + \frac{89}{2^3} + \frac{89}{2^4} + \frac{89}{2^5} + \frac{89}{2^6} = \frac{89}{2} + \frac{89}{4} + \frac{89}{8} + \frac{89}{16} + \frac{89}{32} + \frac{89}{64} = 44 + 22 + 11 + 5 + 2 + 1 = 85\)
• Similarly, instances of 3: \(\frac{89}{3^1} + \frac{89}{3^2} + \frac{89}{3^3} + \frac{89}{3^4} = \frac{89}{3} + \frac{89}{9} + \frac{89}{27} + \frac{89}{81} = 29 + 9 + 3 + 1 = 42\)
Now, the number 12 is formed by two instances of 2 and one instance of 3.
- • From 85 instances of 2's, number of \(2^2\) can be formed \(= \frac{85}{2} = 42\)
• Hence, number of pairs possible of \(2^2\) and 3 = minimum (42, 42) = 42
Hence, the correct answer is option D.
Answer: D
General Discussion
Archit3110
Updated on: 13 Dec 2018, 04:12
Originally posted by Archit3110 on 12 Dec 2018, 03:47.
Last edited by Archit3110 on 13 Dec 2018, 04:12, edited 2 times in total.
Last edited by Archit3110 on 13 Dec 2018, 04:12, edited 2 times in total.
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question is to find factors 89!/12
12= 2^2 * 3
with 2 it would be 85 and 3 it would be 42 so greatest value would be 42
~ 42 IMO D
12= 2^2 * 3
with 2 it would be 85 and 3 it would be 42 so greatest value would be 42
~ 42 IMO D
EgmatQuantExpert
