1986 AIME Problems/Problem 1: Difference between revisions
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This means that <math>\sqrt[4]{x} = y = 3</math> or '''<math>4</math>'''. | This means that <math>\sqrt[4]{x} = y = 3</math> or '''<math>4</math>'''. | ||
Thus the sum of the possible solutions for '''<math>x</math>''' is '''<math>4^4 + 3^4 = 337</math>'''. | Thus the sum of the possible solutions for '''<math>x</math>''' is '''<math>4^4 + 3^4 = \boxed{337}</math>'''. | ||
== See also == | == See also == | ||
Latest revision as of 12:34, 22 July 2020
Problem
What is the sum of the solutions to the equation
?
Solution
Let
. Then we have
, or, by simplifying,
This means that
or
.
Thus the sum of the possible solutions for
is
.
See also
| 1986 AIME (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
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