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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+ | ==Animated Video Solution== | ||
+ | https://youtu.be/zmRiG52jxpg | ||
+ | |||
+ | ~Star League (https://starleague.us) |
Revision as of 18:31, 24 January 2023
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.
Animated Video Solution
~Star League (https://starleague.us)