Difference between revisions of "2021 Fall AMC 10B Problems/Problem 10"
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− | We see that <math>225</math> is one such square that works. Bob gets <math>2</math> and Alice gets <math>25</math> which is valid. Thus, <math>2 + 25 = 27.</math> So <math>\boxed{\texbf(A) }27}</math> is our answer. | + | We see that <math>225</math> is one such square that works. Bob gets <math>2</math> and Alice gets <math>25</math> which is valid. Thus, <math>2 + 25 = 27.</math> So <math>\boxed{\texbf{(A) }27}</math> is our answer. |
-D1r | -D1r |
Revision as of 17:02, 20 October 2023
Contents
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
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)
Solution 2 - Guessing those squares
We see that is one such square that works. Bob gets and Alice gets which is valid. Thus, So $\boxed{\texbf{(A) }27}$ (Error compiling LaTeX. Unknown error_msg) is our answer.
-D1r
Sidenote
Note that Bob's statement (that he knows who won/lost) comes after Alice tells him that she doesn't know who won. Since Alice doesn't know who won, Bob knows that she didn't get draw a (or a ), which tells him that his must be lower than Alice's number.
~jd9
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=1524
Video Solution
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/RyN-fKNtd3A?t=1474
~IceMatrix
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.