1966 AHSME Problems/Problem 5
From AoPSWiki
Problem
The number of values of
satisfying the equation
is:
Solution
Since
is in the denominator,
. Simplifying,
Thus
, which isn't in the domain of the equation. Thus there are no values of
.
See also
| 1966 AHSME (Problems) | ||
| Preceded by Problem 4 | Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||




