Difference between revisions of "2018 AMC 10A Problems/Problem 14"
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==Solution 6== | ==Solution 6== | ||
− | Notice how <math>\frac{3^{100}+2^{100}}{3^{96}+2^{96}}</math> can be rewritten as <math>\frac{81(3^{96})+16(2^{96})}{3^{96}+2^{96}}=\frac{81(3^{96})+81(2^{96})}{3^{96}+2^{96}}-\frac{65(2^{96})}{3^{96}+2^{96}}=81-\frac{65(2^{96})}{3^{96}+2^{96}}</math>. Note that <math>\frac{65(2^96)}{3^{96}+2^{96}}<1</math>, so the greatest integer less than or equal to <math>\frac{3^{100}+2^{100}}{3^{96}+2^{96}}</math> is <math>80</math> or <math>\boxed{\textbf{(A)}}</math> | + | Notice how <math>\frac{3^{100}+2^{100}}{3^{96}+2^{96}}</math> can be rewritten as <math>\frac{81(3^{96})+16(2^{96})}{3^{96}+2^{96}}=\frac{81(3^{96})+81(2^{96})}{3^{96}+2^{96}}-\frac{65(2^{96})}{3^{96}+2^{96}}=81-\frac{65(2^{96})}{3^{96}+2^{96}}</math>. Note that <math>\frac{65(2^{96})}{3^{96}+2^{96}}<1</math>, so the greatest integer less than or equal to <math>\frac{3^{100}+2^{100}}{3^{96}+2^{96}}</math> is <math>80</math> or <math>\boxed{\textbf{(A)}}</math> |
~blitzkrieg21 | ~blitzkrieg21 | ||
{{AMC10 box|year=2018|ab=A|num-b=13|num-a=15}} | {{AMC10 box|year=2018|ab=A|num-b=13|num-a=15}} |
Revision as of 20:22, 9 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 .
Solution 4
A faster solution. Recognize that for exponents of this size will be enormously greater than , so the terms involving will actually have very little effect on the quotient. Now we know the answer will be very close to .
Notice that the terms being added on to the top and bottom are in the ratio with each other, so they must pull the ratio down from 81 very slightly. (In the same way that a new test score lower than your current cumulative grade always must pull that grade downward.) Answer: .
Solution 5 (Using the answer choices)
We can compare the given value to each of our answer choices. We already know that it is greater than because otherwise there would have been a smaller answer, so we move onto . We get:
Cross multiply to get:
Cancel out and divide by to get . We know that , which means the expression is less than so the answer is .
Solution 6
Notice how can be rewritten as . Note that , so the greatest integer less than or equal to is or ~blitzkrieg21
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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All AMC 10 Problems and Solutions |