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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 10"

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<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>
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== Solution 1 ==
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Let Alice have the number A, Bob B.
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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.
  
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== Solution==
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--stjwyl
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== Solution 2==
 
Denote by <math>A</math> and <math>B</math> the numbers drawn by Alice and Bob, respectively.
 
Denote by <math>A</math> and <math>B</math> the numbers drawn by Alice and Bob, respectively.
  
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~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
  
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== Solution 2 - Guessing those Squares ==  
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== 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.
 
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.

Latest revision as of 16:21, 8 September 2024

Problem

Forty slips of paper numbered $1$ to $40$ 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 $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?

$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$

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 $A$ cannot equal $1$. Therefore, it makes sense that Bob has $2$ because he now knows that Alice has the larger number. $2$ is also prime. The last statement means that $200+A$ is a perfect square. The three squares in the range $200-300$ are $225$, $256$, and $289$. So, $A$ could equal $25$, $56$, or $89$, so $A+B$ is $27$, $58$, or $91$, of only $\boxed{\textbf{(A) }27}$ is an answer choice.

--stjwyl

Solution 2

Denote by $A$ and $B$ the numbers drawn by Alice and Bob, respectively.

Alice's sentence “I can't tell who has the larger number.” implies $A \in \left\{ 2 , \cdots , 39 \right\}$.

Bob's sentence “I know who has the larger number.” implies $B \in \left\{ 1 , 2 , 39, 40 \right\}$.

Their subsequent conversation that $B$ is prime implies $B = 2$.

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 $200 + A$ is a perfect square. Hence, $A = 25$.

Therefore, the answer is $\boxed{\textbf{(A) }27}$.

~Steven Chen (www.professorchenedu.com)

Solution 3 - Guessing those Squares

We see that $225$ is one such square that works. Bob gets $2$ and Alice gets $25$ which is valid. Thus, $2 + 25 = 27.$ So $\boxed{\textbf{(A) }27}$ 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 $1$ (or a $40$), which tells him that his $2$ 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

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

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