Difference between revisions of "2023 AMC 8 Problems/Problem 18"
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We have <math>2</math> directions going <math>5</math> right or <math>3</math> left. We can assign a variable to each of these directions. We can call going right 1 direction X and we can call going 1 left Y. We can build a equation of <math>5X-3Y=2023</math>. Where we have to limit the number of moves we do. We can do this by making more of our moves the <math>5</math> move turn then the <math>3</math> move turn. The first obvious step is to go some amount of moves in the → direction then subtract off in the ← direction to land on <math>2023</math>. The least amount of <math>3</math>’s added to <math>2023</math> to make a multiple of <math>5</math> is <math>4</math> as <math>2023 + 4(3) = 2035</math>. So now we have solved the problem as we just go <math>\frac{2035}{5} = 407</math> hops right, and just do 4 more hops left. Yielding <math>407 + 4 = \boxed{\text{(D)}411}</math> as our answer. | We have <math>2</math> directions going <math>5</math> right or <math>3</math> left. We can assign a variable to each of these directions. We can call going right 1 direction X and we can call going 1 left Y. We can build a equation of <math>5X-3Y=2023</math>. Where we have to limit the number of moves we do. We can do this by making more of our moves the <math>5</math> move turn then the <math>3</math> move turn. The first obvious step is to go some amount of moves in the → direction then subtract off in the ← direction to land on <math>2023</math>. The least amount of <math>3</math>’s added to <math>2023</math> to make a multiple of <math>5</math> is <math>4</math> as <math>2023 + 4(3) = 2035</math>. So now we have solved the problem as we just go <math>\frac{2035}{5} = 407</math> hops right, and just do 4 more hops left. Yielding <math>407 + 4 = \boxed{\text{(D)}411}</math> as our answer. | ||
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+ | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat | ||
==Animated Video Solution== | ==Animated Video Solution== | ||
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~Star League (https://starleague.us) | ~Star League (https://starleague.us) | ||
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Revision as of 19:57, 24 January 2023
Written Solution
We have directions going right or left. We can assign a variable to each of these directions. We can call going right 1 direction X and we can call going 1 left Y. We can build a equation of . Where we have to limit the number of moves we do. We can do this by making more of our moves the move turn then the move turn. The first obvious step is to go some amount of moves in the → direction then subtract off in the ← direction to land on . The least amount of ’s added to to make a multiple of is as . So now we have solved the problem as we just go hops right, and just do 4 more hops left. Yielding as our answer.
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
Animated Video Solution
~Star League (https://starleague.us)