Difference between revisions of "2018 AMC 10A Problems/Problem 14"
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<cmath>3^{100}+2^{100}=x(3^{96}+2^{96}).</cmath> | <cmath>3^{100}+2^{100}=x(3^{96}+2^{96}).</cmath> | ||
Now let's take a look at the answer choices. We notice that <math>81</math>, choice <math>B</math>, can be written as 3^4. Plugging this into out equation above, we get | Now let's take a look at the answer choices. We notice that <math>81</math>, choice <math>B</math>, can be written as 3^4. Plugging this into out equation above, we get | ||
− | <cmath>3^{100}+2^{100} ? 3^4(3^{96}+2^{96}) \Rightarrow 3^{100}+2^{100} ? 3^{100}+3^4*2^{96}.</cmath> | + | <cmath>3^{100}+2^{100} ?=? 3^4(3^{96}+2^{96}) \Rightarrow 3^{100}+2^{100} ?=? 3^{100}+3^4*2^{96}.</cmath> |
The right side is larger than the left side because | The right side is larger than the left side because | ||
<cmath>2^{100} \leq 2^{96}*3^4.</cmath> | <cmath>2^{100} \leq 2^{96}*3^4.</cmath> | ||
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>. |
Revision as of 14:42, 8 February 2018
What is the greatest integer less than or equal to
Solution
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 3^4. 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 .