Mind Blowing exponents
It is given that and , where is a real number and is the largest integer smaller than or equal to .
Find the number of possible values of .
Note: This question is taken from The Hong Kong Youth Mathematical Challenge 2014.
The answer is 47.
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1 solution
We have (1): 1 1 ≤ y 2 < 1 2 and (2): 4 1 ≤ y 3 < 4 2 .
(1) 3 gives (3): 1 3 3 1 ≤ y 6 < 1 7 2 8 .
(2) 2 gives (4): 1 6 8 1 ≤ y 6 < 1 7 6 4 .
Combining (3) and (4): 1 6 8 1 ≤ y 6 < 1 7 2 8 .
Therefore the number of possible values is 1 7 2 8 − 1 6 8 1 = 4 7 .
Note: This solution is not taken from The Hong Kong Youth Mathematical Challenge 2014.