Difference between revisions of "2021 Fall AMC 10B Problems/Problem 10"
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==Video Solution by Interstigation== | ==Video Solution by Interstigation== |
Revision as of 17:52, 16 August 2022
Problem
Forty slips of paper numbered to are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?
Solution 1
Because Alice doesn't know who has the larger number, she doesn't have Because Alice says that she doesn't know who has the larger number, Bob knows that she doesn't have But Bob knows who has the larger number, this implies that Bob has the smallest possible number. Because Bob's number is prime, Bob's number is . Thus, the perfect square is in the The only perfect square is Thus, Alice's number is The sum of Alice's and Bob's number is Thus the answer is
~NH14
Solution 2
Denote by and the numbers drawn by Alice and Bob, respectively.
Alice's sentence ``I can't tell who has the larger number. implies .
Bob's sentence ``I know who has the larger number. implies .
Their subsequent conversation that is prime implies .
Then, Alice's next sentence ``In that case, if I multiply your number by 100 and add my number, the result is a perfect square. implies is a perfect square. Hence, .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=1524
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.