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03 Jul 2020, 05:35
Kudos
Bookmarks
D
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:
33% (01:25) correct
67%
(01:20)
wrong
based on 176
sessions
History
Date
Time
Result
Not Attempted Yet
If \(x^y = 64\) and x and y are integers, then x + y can take how many values
A. 3
B. 4
C. 5
D. 6
E. 7
A. 3
B. 4
C. 5
D. 6
E. 7
03 Jul 2020, 05:41
Kudos
Bookmarks
Explanation:
x & y are integers
x^y = 64
1. 64^1 = 64+1 = 65
2. 2^6 = 2+6 = 8
3. (-2)^6 = -2+6 = 4
4. 4^3 = 4+3 = 7
5. 8^2 = 8+2 = 10
6. (-8)^2 = -8+2 = 6
6 values.
IMO-D
x & y are integers
x^y = 64
1. 64^1 = 64+1 = 65
2. 2^6 = 2+6 = 8
3. (-2)^6 = -2+6 = 4
4. 4^3 = 4+3 = 7
5. 8^2 = 8+2 = 10
6. (-8)^2 = -8+2 = 6
6 values.
IMO-D
03 Jul 2020, 05:43
Kudos
Bookmarks
Bunuel
\(x^y = 64 = 2^6 = (-2)^6 = 4^3 = 64^1 = 8^2 = (-8)^2\)
Six possible values
Answer: Option D
