Difference between revisions of "2021 Fall AMC 10B Problems/Problem 10"
(→Solution) |
m (→Solution 1) |
||
(24 intermediate revisions by 14 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | + | Forty slips of paper numbered <math>1</math> to <math>40</math> 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 <math>100</math> and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat? | |
<math>\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67</math> | <math>\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67</math> | ||
+ | == Solution 1 == | ||
+ | Let Alice have the number A, Bob B. | ||
+ | When Alice says that she can't tell who has the larger number, it means that <math>A</math> cannot equal <math>1</math>. Therefore, it makes sense that Bob has <math>2</math> because he now knows that Alice has the larger number. <math>2</math> is also prime. The last statement means that <math>200+A</math> is a perfect square. The three squares in the range <math>200-300</math> are <math>225</math>, <math>256</math>, and <math>289</math>. So, <math>A</math> could equal <math>25</math>, <math>56</math>, or <math>89</math>, so <math>A+B</math> is <math>27</math>, <math>58</math>, or <math>91</math>, of only <math>\boxed{\textbf{(A) }27}</math> is an answer choice. | ||
− | + | --stjwyl | |
− | |||
− | + | == Solution 2== | |
+ | Denote by <math>A</math> and <math>B</math> the numbers drawn by Alice and Bob, respectively. | ||
+ | Alice's sentence “I can't tell who has the larger number.” implies <math>A \in \left\{ 2 , \cdots , 39 \right\}</math>. | ||
+ | |||
+ | Bob's sentence “I know who has the larger number.” implies <math>B \in \left\{ 1 , 2 , 39, 40 \right\}</math>. | ||
+ | |||
+ | Their subsequent conversation that <math>B</math> is prime implies <math>B = 2</math>. | ||
+ | |||
+ | 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 <math>200 + A</math> is a perfect square. | ||
+ | Hence, <math>A = 25</math>. | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(A) }27}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | == Solution 3 - Guessing those Squares == | ||
+ | |||
+ | 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{\textbf{(A) }27}</math> 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 <math>1</math> (or a <math>40</math>), which tells him that his <math>2</math> must be lower than Alice's number. | ||
+ | |||
+ | ~jd9 | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/p9_RH4s-kBA?t=1524 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/vB3R4b-X3yA | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/j-AZyR78-Ns | ||
+ | |||
+ | ~savannahsolver | ||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/RyN-fKNtd3A?t=1474 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=11|num-b=9}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=11|num-b=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:21, 8 September 2024
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 1
Let Alice have the number A, Bob B. When Alice says that she can't tell who has the larger number, it means that cannot equal . Therefore, it makes sense that Bob has because he now knows that Alice has the larger number. is also prime. The last statement means that is a perfect square. The three squares in the range are , , and . So, could equal , , or , so is , , or , of only is an answer choice.
--stjwyl
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)
Solution 3 - Guessing those Squares
We see that is one such square that works. Bob gets and Alice gets which is valid. Thus, So 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.