Difference between revisions of "2018 AMC 10A Problems/Problem 14"
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Let's set this value equal to <math>x</math>. We can write | Let's set this value equal to <math>x</math>. We can write | ||
<cmath>\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=x.</cmath> | <cmath>\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=x.</cmath> |
Revision as of 21:52, 8 February 2018
What is the greatest integer less than or equal to
Contents
Solution
Solution 1
Let's set this value equal to . We can write Multiplying by on both sides, we get Now let's take a look at the answer choices. We notice that , choice , can be written as . Plugging this into out equation above, we get The right side is larger than the left side because This means that our original value, , must be less than . The only answer that is less than is so our answer is .
~Nivek
Solution 2
Let and . Then our fraction can be written as . Notice that . So , . And our only answer choice less than 81 is
~RegularHexagon
Solution 3
Let . Multiply both sides by , and expand. Rearranging the terms, we get . The left side is strictly decreasing, and it is negative when . This means that the answer must be less than ; therefore the answer is .
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 | ||
All AMC 10 Problems and Solutions |