Difference between revisions of "2018 AMC 10B Problems/Problem 20"
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<math>\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}</math> | <math>\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}</math> | ||
− | ==Solution 1 | + | ==Solution 1 (Elegant and Fast)== |
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− | < | + | We know <cmath>f(3) = f(2) - f(1) + 3</cmath> <cmath>f(4) = f(3) - f(2) + 4</cmath> <cmath>f(5) = f(4) - f(3) + 5</cmath> <cmath>\vdots</cmath> <cmath>f(2018) = f(2017) - f(2016) + 2018</cmath> Seeing all those negatives remind us of telescoping sums. This suggests we add the equations. Adding we get <cmath>f(3) + f(4) + f(5) + \cdots + f(2018) = 3 + 4 + 5 + \cdots + 2018 - f(1) = 3 + 4 + 5 + \cdots + 2017 = \frac{(2017)(2018)}{2} - 3.</cmath> |
− | < | + | Notice that <cmath>f(2018) = f(3) + f(4) + f(5) + \cdots + f(2018) - [f(3) + f(4) + f(5) + \cdots + f(2017)].</cmath> Using the same telescoping logic we see that <cmath>f(3) + f(4) + f(5) + \cdots + f(2017) = \frac{(2016)(2017)}{2} -3.</cmath> So subtracting <cmath>\frac{(2017)(2018)}{2} - 3 - \left(\frac{(2016)(2017)}{2} - 3\right) = 2017.</cmath> This simply means the answer is <math>\boxed{\textbf{(B)} \text{ 2017}}.</math> |
− | + | ~coolmath_2018 | |
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==Solution 2 (A Bit Bashy)== | ==Solution 2 (A Bit Bashy)== | ||
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<cmath>f(15)=15</cmath> | <cmath>f(15)=15</cmath> | ||
<cmath>.....</cmath> | <cmath>.....</cmath> | ||
− | Notice that <math>f(n)=n</math> whenever <math>n</math> is an odd multiple of <math>3</math>, and the pattern of numbers that follow will always be +3, +2, +0, -1, +0. | + | Notice that <math>f(n)=n</math> whenever <math>n</math> is an odd multiple of <math>3</math>, and the pattern of numbers that follow will always be <math>+3</math>, <math>+2</math>, <math>+0</math>, <math>-1</math>, <math>+0</math>. |
The largest odd multiple of <math>3</math> smaller than <math>2018</math> is <math>2013</math>, so we have | The largest odd multiple of <math>3</math> smaller than <math>2018</math> is <math>2013</math>, so we have | ||
<cmath>f(2013)=2013</cmath> | <cmath>f(2013)=2013</cmath> | ||
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==Solution 3 (Bashy Pattern Finding)== | ==Solution 3 (Bashy Pattern Finding)== | ||
Writing out the first few values, we get: | Writing out the first few values, we get: | ||
− | <math>1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19...</math>. Examining, we see that every number <math>x</math> where <math>x \equiv 1 \pmod 6</math> has <math>f(x)=x</math>, <math>f(x+1)=f(x)=x</math>, and <math>f(x-1)=f(x-2)=x+1</math>. The greatest number that's 1 | + | <math>1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19...</math>. Examining, we see that every number <math>x</math> where <math>x \equiv 1\pmod 6</math> has <math>f(x)=x</math>, <math>f(x+1)=f(x)=x</math>, and <math>f(x-1)=f(x-2)=x+1</math>. The greatest number that's <math>1\pmod{6}</math> and less <math>2018</math> is <math>2017</math>, so we have <math>f(2017)=f(2018)=2017.</math> <math>\boxed B</math> |
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==Solution 4 (Algebra)== | ==Solution 4 (Algebra)== | ||
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After plugging in <math>n-3</math> to the equation above and doing some algebra, we have that <math>f(n)-f(n-6)=6</math>. | After plugging in <math>n-3</math> to the equation above and doing some algebra, we have that <math>f(n)-f(n-6)=6</math>. | ||
Consequently, | Consequently, | ||
− | <cmath>f(2018)-f(2012)=6</cmath> | + | <cmath>f(2018)-f(2012)=6.</cmath> |
− | <cmath>f(2012)-f(2006)=6</cmath> | + | <cmath>f(2012)-f(2006)=6.</cmath> |
<cmath>\ldots</cmath> | <cmath>\ldots</cmath> | ||
− | <cmath>f(8)-f(2)=6</cmath> | + | <cmath>f(8)-f(2)=6.</cmath> |
− | Adding these <math>336</math> equations up, we have that <math>f(2018)-f(2)=6 | + | Adding these <math>336</math> equations up, we have that <math>f(2018)-f(2)=6 \cdot 336</math> and <math>f(2018)=\boxed{2017}</math>. |
~AopsUser101 | ~AopsUser101 | ||
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==See Also== | ==See Also== |
Revision as of 23:58, 25 August 2020
Contents
Problem
A function is defined recursively by and for all integers . What is ?
Solution 1 (Elegant and Fast)
We know Seeing all those negatives remind us of telescoping sums. This suggests we add the equations. Adding we get
Notice that Using the same telescoping logic we see that So subtracting This simply means the answer is
~coolmath_2018
Solution 2 (A Bit Bashy)
Start out by listing some terms of the sequence.
Notice that whenever is an odd multiple of , and the pattern of numbers that follow will always be , , , , . The largest odd multiple of smaller than is , so we have
Solution 3 (Bashy Pattern Finding)
Writing out the first few values, we get: . Examining, we see that every number where has , , and . The greatest number that's and less is , so we have
Solution 4 (Algebra)
Adding the two equations, we have that Hence, . After plugging in to the equation above and doing some algebra, we have that . Consequently, Adding these equations up, we have that and .
~AopsUser101
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.