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02 Nov 2019, 10:28
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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:
95%
(hard)
Question Stats:
31% (01:51) correct
69%
(02:00)
wrong
based on 115
sessions
History
Date
Time
Result
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How many rational numbers in between 0 and 1 are such that, when expressed as a fraction in the lowest form, the product of their numerator and denominator is 20! ?
A. 4
B. 32
C. 96
D. 128
E. 256
A. 4
B. 32
C. 96
D. 128
E. 256
02 Nov 2019, 19:29
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nick1816
To answer this, you should know..
(a) 20! = 1*2*3....19*20
(b) Rational number are all numbers that can be written as a fraction
(c) For a number to be between 0 and 1, the numerator should be less than the denominator
Now the number is \(\frac{a}{b}\), where a<b, and these a and b are co-prime, that is no common factors between them and should contain all numbers from 2 to 20 within themselves.
This means they cannot have SAME prime number in both and they have to contain all prime numbers 2, 3, 5, 7, 11, 13, 17, and 19 with some power.
So, if 2 is there in a, it will contain all 2s that are in 20!.
Now, the solution becomes simple -- that is the numerator will either have a prime number or not.
There are 8 prime numbers - 2, 3, 5, 7, 11, 13, 17 and 19.
Let us just take the numerator, it can have none, some or all of the 8 numbers in \(2*2*2*2*2*2*2*2=2^8\) ways.
Now there can be \(2^8\) such numerators but half of them will have numerator>denominator and the other half will have denominator>numerator.
Thus our answer for numerator<denominator = \(\frac{2^8}{2}=2^7=128\)
D
