Difference between revisions of "2021 AMC 10B Problems/Problem 17"
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\end{align*}</cmath> | \end{align*}</cmath> | ||
Therefore, the answer is <math>\boxed{\textbf{(C) }\text{Ravon was given card 4.}}</math> | Therefore, the answer is <math>\boxed{\textbf{(C) }\text{Ravon was given card 4.}}</math> | ||
− | + | Note that if we didn't notice 4 can be composed of only 1 and 3, we can do casework from Ravon or Tyrone (only one of them can have the card numbered 10) and arrive at the same conclusions. | |
Certainly, if we read the answer choices sooner, then we can stop at <math>(\bigstar)</math> and pick <math>\textbf{(C)}.</math> | Certainly, if we read the answer choices sooner, then we can stop at <math>(\bigstar)</math> and pick <math>\textbf{(C)}.</math> | ||
Revision as of 12:16, 19 October 2021
Contents
[hide]Problem
Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given cards out of a set of cards numbered The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon-- Oscar-- Aditi-- Tyrone-- Kim-- Which of the following statements is true?
Solution
By logical deduction, we consider the scores from lowest to highest: Therefore, the answer is Note that if we didn't notice 4 can be composed of only 1 and 3, we can do casework from Ravon or Tyrone (only one of them can have the card numbered 10) and arrive at the same conclusions. Certainly, if we read the answer choices sooner, then we can stop at and pick
~smarty101 ~smartypantsno_3 ~SmileKat32 ~MRENTHUSIASM
Video Solution by OmegaLearn (Using Logical Deduction)
~ pi_is_3.14
Video Solution by TheBeautyofMath
https://youtu.be/FV9AnyERgJQ?t=284
~IceMatrix
Video Solution by Interstigation
~Interstigation
See Also
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.