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. | |
Revision as of 18:26, 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.
