Difference between revisions of "2018 AMC 10A Problems/Problem 14"
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This means that our original value, <math>x</math>, must be less than <math>81</math>. The only answer that is less than <math>81</math> is <math>80</math> so our answer is <math>\boxed{A}</math>. | This means that our original value, <math>x</math>, must be less than <math>81</math>. The only answer that is less than <math>81</math> is <math>80</math> so our answer is <math>\boxed{A}</math>. | ||
− | ~Nivek | + | ~Nivek |
==Solution 3== | ==Solution 3== | ||
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Notice that the terms being added on to the top and bottom are in the ratio <math>\frac{1}{16}</math> 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: <math>\boxed{(A)}</math>. | Notice that the terms being added on to the top and bottom are in the ratio <math>\frac{1}{16}</math> 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: <math>\boxed{(A)}</math>. | ||
− | ==Solution 7 (Using the answer choices)== | + | ==Solution 7== |
+ | 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 | ||
+ | |||
+ | ==Solution 8== | ||
+ | For positive <math>a, b, c, d</math>, if <math>\frac{a}{b}<\frac{c}{d}</math> then <math>\frac{c+a}{d+b}<\frac{c}{d}</math>. Let <math>a=2^{100}, b=2^{96}, c=3^{100}, d=3^{96}</math>. Then <math>\frac{c}{d}=3^4</math>. So answer is less than 81, which leaves only one choice, 80. | ||
+ | * Note that the algebra here is synonymous to the explanation given in Solution 6. This is the algebraic reason to the logic of if you get a test score with a lower percentage than your average (no matter how many points/percentage of your total grade it was worth), it will pull your overall grade down. | ||
+ | |||
+ | ~ ccx09 | ||
+ | |||
+ | ==Solution 9== | ||
+ | Try long division, and notice putting <math>3^4=81</math> as the denominator is too big and putting <math>3^4-1=80</math> is too small. So we know that the answer is between <math>80</math> and <math>81</math>, yielding <math>80</math> as our answer. | ||
+ | |||
+ | ==Solution 10 (Using the answer choices)== | ||
+ | ===Solution 10.1=== | ||
+ | |||
We can compare the given value to each of our answer choices. We already know that it is greater than <math>80</math> because otherwise there would have been a smaller answer, so we move onto <math>81</math>. We get: | We can compare the given value to each of our answer choices. We already know that it is greater than <math>80</math> because otherwise there would have been a smaller answer, so we move onto <math>81</math>. We get: | ||
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Cancel out <math>3^{100}</math> and divide by <math>2^{96}</math> to get <math>2^{4} \text{ ? }3^4</math>. We know that <math>2^4 < 3^4</math>, which means the expression is less than <math>81</math> so the answer is <math>\boxed{(A)}</math>. | Cancel out <math>3^{100}</math> and divide by <math>2^{96}</math> to get <math>2^{4} \text{ ? }3^4</math>. We know that <math>2^4 < 3^4</math>, which means the expression is less than <math>81</math> so the answer is <math>\boxed{(A)}</math>. | ||
− | ==Solution | + | ===Solution 10.2=== |
− | + | ||
− | + | We know this will be between 16 and 81 because <math>\frac{3^{100}}{3^{96}} = 3^4 = 81</math> and <math>\frac{2^{100}}{2^{96}} = 2^4 = 16</math>. <math>80=\boxed{(A)}</math> is the only option choice in this range. | |
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− | |||
− | |||
− | + | ==Explanation for why 80 is indeed the floor== | |
− | == | + | We need <math>3^{100}+2^{100} > 80 \cdot 3^{96} + 5 \cdot 2^{100}</math>. Since <math>3^{100} = 81\cdot 3^{96}</math>, this translates to |
− | + | <cmath>3^{96} > 4\cdot 2^{100} = 64\cdot 2^{96}.</cmath> | |
+ | We now prove that <math>(3/2)^k > k</math> for all positive integers <math>k</math>. | ||
+ | Clearly, <math>(3/2)^2 = 2.25 > 2</math>. Assume <math>(3/2)^k > k</math> where <math>k\ge 2</math>. Then <math>\left(\frac{3}{2}\right)^{k+1} > \frac{3k}{2} = k + \frac{k}{2}</math>. But since <math>k/2 \ge 1</math>, we have that <math>(3/2)^{k+1} > k+1</math>. By induction (and <math>k=1</math> is trivial), the claim is proven. | ||
− | + | Thus, <math>\left(\frac{3}{2}\right)^{96} > 96 > 64</math>. Writing this proof backwards and dividing both sides of the initial equation by <math>80</math> yields <math>80 < \frac{3^{100}+2^{100}}{3^{96}+2^{96}} < 81</math>. | |
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==See Also== | ==See Also== |
Latest revision as of 19:58, 28 December 2020
Contents
Problem
What is the greatest integer less than or equal to
Solution 1
We write Hence we see that our number is a weighted average of 81 and 16, extremely heavily weighted toward 81. Hence the number is ever so slightly less than 81, so the answer is .
Solution 2
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 our 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 3
.
We can ignore the 1's on the end because they won't really affect the fraction. So, the answer is very very very close but less than the new fraction.
So, our final answer is very close but not quite 81, and therefore the greatest integer less than the number is
Solution 4
Let and . Then our fraction can be written as . Notice that . So , . And our only answer choice less than 81 is (RegularHexagon)
Solution 5
Let . Multiply both sides by , and expand. Rearranging the terms, we get . The left side is decreasing, and it is negative when . This means that the answer must be less than ; therefore the answer is .
Solution 6 (eyeball it)
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 7
Notice how can be rewritten as . Note that , so the greatest integer less than or equal to is or ~blitzkrieg21
Solution 8
For positive , if then . Let . Then . So answer is less than 81, which leaves only one choice, 80.
- Note that the algebra here is synonymous to the explanation given in Solution 6. This is the algebraic reason to the logic of if you get a test score with a lower percentage than your average (no matter how many points/percentage of your total grade it was worth), it will pull your overall grade down.
~ ccx09
Solution 9
Try long division, and notice putting as the denominator is too big and putting is too small. So we know that the answer is between and , yielding as our answer.
Solution 10 (Using the answer choices)
Solution 10.1
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 10.2
We know this will be between 16 and 81 because and . is the only option choice in this range.
Explanation for why 80 is indeed the floor
We need . Since , this translates to We now prove that for all positive integers . Clearly, . Assume where . Then . But since , we have that . By induction (and is trivial), the claim is proven.
Thus, . Writing this proof backwards and dividing both sides of the initial equation by yields .
See Also
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 |
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