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Difference between revisions of "2021 JMPSC Problems/Problem 2"
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== Problem ==
 
== Problem ==
  
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Brady has an unlimited supply of quarters (<math>0.25), dimes (</math>0.10), nickels (<math>0.05), and pennies (</math>0.01). What is the least number (quantity, not type) of coins Brady can use to pay off <math>\text{</math>}2.78<math>?
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Brady has an unlimited supply of quarters (<math>0.25), dimes (</math>0.10), nickels (<math>0.05), and pennies (</math>0.01). What is the least number (quantity, not type) of coins Brady can use to pay off [bbtext]<math>[/bbtext]</math>2.78<math>?
  
 
== Solution ==
 
== Solution ==
  
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To minimize the number of coins, we need to maximize the amount of high-valued coins we use. 11 quarters are worth </math>\text{<math>}2.75</math>, which is the most quarters we can use to get a value less than or equal to <math>\text{</math>}2.78<math>. Finally, we can add 3 pennies to get a total of </math>\text{<math>}2.87</math>, so the answer is <math>11+3=14</math>.
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To minimize the number of coins, we need to maximize the amount of high-valued coins we use. 11 quarters are worth [bbtext]</math>[/bbtext]<math>2.75</math>, which is the most quarters we can use to get a value less than or equal to [bbtext]<math>[/bbtext]</math>2.78<math>. Finally, we can add 3 pennies to get a total of [bbtext]</math>[/bbtext]<math>2.87</math>, so the answer is <math>11+3=14</math>.

Revision as of 18:24, 10 July 2021

Problem

Brady has an unlimited supply of quarters ($0.25), dimes ($0.10), nickels ($0.05), and pennies ($0.01). What is the least number (quantity, not type) of coins Brady can use to pay off [bbtext]$[/bbtext]$2.78$?

== Solution ==

To minimize the number of coins, we need to maximize the amount of high-valued coins we use. 11 quarters are worth [bbtext]$ (Error compiling LaTeX. Unknown error_msg)[/bbtext]$2.75$, which is the most quarters we can use to get a value less than or equal to [bbtext]$[/bbtext]$2.78$. Finally, we can add 3 pennies to get a total of [bbtext]$[/bbtext]$2.87$, so the answer is $11+3=14$.

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