Logical guessing game!

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Mathdreams

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Mathdreams

#1 h Nov 17, 2021, 4:09 PM • 2 Y

Y by HWenslawski, megarnie

Fourty 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?

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

This post has been edited 1 time. Last edited by Mathdreams, Nov 17, 2021, 4:15 PM

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Toinfinity

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Toinfinity

#2 h Nov 17, 2021, 4:10 PM • 1 Y

Y by wamofan

Was it (A) 27, 2 and 25

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Mathlete12345654

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Mathlete12345654

#3 h Nov 17, 2021, 4:11 PM

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Can confirm @above

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mathboy100

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mathboy100

#4 h Nov 17, 2021, 4:18 PM

Y by

Mathdreams wrote:

Fourty 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?

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

If Alice does not know who has the largest number, then she doesn't have $1$ or $40$ . Thus, Bob must have $2$ or $39$ . $2$ is prime, so Bob has $2$ and Alice has $25$ . The answer is $27$ .

This post has been edited 1 time. Last edited by mathboy100, Nov 17, 2021, 4:18 PM

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mewto

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mewto

#5 h Nov 17, 2021, 4:20 PM

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Yes, I got A. I'm so relieved

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amopuri

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amopuri

#6 h Nov 17, 2021, 4:35 PM

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oops i kept thinking that bob had to have a bigger number so i kept getting stuck with $39$

skipped it, came back to it, and got it tho

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StopSine

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StopSine

#7 h Nov 17, 2021, 4:46 PM

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Why did I think that 225 was not a perfect square for the first 3 minutes of the problem.

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buddy2007

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buddy2007

#8 h Nov 17, 2021, 5:00 PM

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Mathdreams wrote:

Fourty 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?

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

I got 2 answers, 27 (225) and 67 (3136) but i had to use a bit of logic to get (A)

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IceMatrix

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IceMatrix

#9 h Nov 17, 2021, 5:09 PM • 1 Y

Y by FIREDRAGONMATH16

I liked this one. Thought it was fairly standard logic application for experienced solvers, and the solution process felt satisfying. Those uninitiated(first or second time test takers) probably had more difficulty though.
If you are curious: https://youtu.be/RyN-fKNtd3A?t=1474

This post has been edited 1 time. Last edited by IceMatrix, Nov 17, 2021, 5:12 PM
Reason: Added link

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jasperE3

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jasperE3

#10 h Nov 17, 2021, 5:23 PM • 1 Y

Y by megarnie

I agree with #9.
For Alice to not know who has the larger number, she must not have either $1$ or $40$ . Thus, Bob needs to have either $1,2,39$ , or $40$ . The only prime out of these is $2$ . Alice's number is $x$ , then $200+x$ is a square. The only such number from $1$ to $40$ is $25$ , so the sum is $2+25=\boxed{\textbf{(A)}~27}$ .

This post has been edited 1 time. Last edited by jasperE3, Nov 17, 2021, 5:23 PM

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HighWater

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HighWater

#11 h Nov 18, 2021, 1:10 AM

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mewto wrote:

Yes, I got A. I'm so relieved

*PHEW!* I talked to a lot of people and some got E, so I was worried I got it wrong, but then I explained my logic and they immediately realized their mistakes.

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asimov

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asimov

#12 h Nov 18, 2021, 1:35 AM

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Can confirm A. It was a very nice and fun problem.

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megarnie

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megarnie

#13 h Nov 18, 2021, 1:55 AM

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We claim that Bob gets a $1$ , $2$ , $39$ , or $40$ .
Proof: If not, then Alice could've gotten a number from either side of Bob.

Since Bob got a prime, Bob got a $2$ .

So Alice got a $25,56,89$ , but $25$ is the only one less than $40$ .

Thus, the answer is $2+25=\boxed{\textbf{(A)}\ 57}$ .

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Overlord123

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Overlord123

#14 h Nov 18, 2021, 2:06 AM

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Solution

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exp-ipi-1

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exp-ipi-1

#15 h Nov 22, 2021, 10:01 PM

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got it wrong because i forgot that 2 was a prime :wacko:

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pog

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pog

#16 h Nov 23, 2021, 1:48 AM

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My bestie DeToasty3 had a lot of fun looking at the solution for this problem

Solution

This post has been edited 2 times. Last edited by pog, Jan 8, 2022, 1:20 AM

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pog

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pog

#17 h Jan 8, 2022, 1:16 AM • 2 Y

Y by HrishiP, DeToasty3

Should say "Forty" and not "Fourty" lol

$\underbrace{2 + 2 + \cdots + 2 + 2}_{100\text{ times}}$

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hansenhe

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hansenhe

#18 h Jan 8, 2022, 1:23 AM

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Post #13 by megarnie

@megarnie 2+25=27...

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DeToasty3

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DeToasty3

#19 h Jan 8, 2022, 1:49 AM • 19 Y

Y by fuzimiao2013, pog, pandabearcat, centslordm, bissue, nikenissan, v4913, mahaler, RedFlame2112, megarnie, RP3.1415, asdf334, HrishiP, math31415926535, rayfish, DankBasher619, john0512, Bryan013, Jack_w

We start with a miraculous claim which is the pith of this problem:

Claim 1. Bob's number must be $2$ .

Proof. Before we can get our hands dirty, we shall prove a few subclaims:

Subclaim 1.1. Alice cannot have the number $1$ .

Proof. If she did, then Alice would know that Bob's number is greater than hers.

Subclaim 1.2. Alice cannot have the number $40$ .

Proof. If she did, then Alice would know that Bob's number is less than hers.

Subclaim 1.3. $1$ is a possibility for Bob's number.

Proof. If $1$ is Bob's number, then he knows that Alice's number is larger than his.

Subclaim 1.4. $2$ is a possibility for Bob's number.

Proof. If $2$ is Bob's number, then he knows that, because Alice's number cannot be $1$ , Alice's number is larger than his.

Subclaim 1.5. $39$ is a possibility for Bob's number.

Proof. If $39$ is Bob's number, then he knows that, because Alice's number cannot be $40$ , Alice's number is less than his.

Subclaim 1.6. $40$ is a possibility for Bob's number.

Proof. If $40$ is Bob's number, then he knows that Alice's number is less than his.

We have deduced that Bob's number can be either $1$ , $2$ , $39$ , or $40$ . Of these four numbers, only $2$ is prime, hence we have proved the claim. $\square$

Now that we know that Bob's number is $2$ , we may use Alice's final sentence to finish the problem. Multiplying $2$ by $100$ is merely equivalent to computing the expression $\underbrace{2 + 2 + 2 + \cdots + 2 + 2}_{100\text{ times}}$ , from which we produce the number $200$ . Now, we must search for perfect squares. Before we do this, we have to bound the range of possible values of this perfect square. To do this, we introduce another claim:

Claim 2. The perfect square must be contained in the interval $(202,240)$ .

To prove this claim, we introduce two subclaims.

Subclaim 2.1. The perfect square must be strictly greater than $202$ .

Proof. As Alice cannot have the number $1$ , she must have at least the number $2$ . However, since Bob has the number $2$ , Alice must have at least the number $3$ . Thus, the perfect square must be at least $200+3=203$ , hence proven.

Subclaim 2.2. The perfect square must be strictly less than $240$ .

Proof. As Alice cannot have the number $40$ , she must have at most the number $39$ . Thus, the perfect square must be at most $200+39=239$ , hence proven.

Combining these two subclaims, we have that the perfect square must be contained in the interval $(202,240)$ , as desired. $\square$

Now, we must search for perfect squares in the interval $(202,240)$ . Upon squaring $15$ , we find that the number obtained is $225$ , which, by sheer luck, is in the interval $(202,240)$ . However, upon squaring $14$ and $16$ , we obtain the numbers $196$ and $256$ , neither of which are in the interval. It is well-known that $f(x)=x^2$ is strictly increasing as $x$ increases from $0$ . Thus, if $14^2$ and $16^2$ are out of the interval $(202,240)$ from opposite sides of the interval, we may conclude that $15^2=225$ is the only possible perfect square. This means that Alice's number is $225-200=25$ .

Finally, we may take the sum of Alice's and Bob's numbers, which are $25$ and $2$ , respectively, for a final answer of $25+2=\boxed{\textbf{(A) }27}$ , and we are done. $\blacksquare$

Remark. This problem is clearly unsuitable for the AMC 10. Proving even one of these claims by itself proves to be a significant challenge, perhaps too much for tenth graders.

This post has been edited 2 times. Last edited by DeToasty3, Jan 8, 2022, 2:19 AM

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fuzimiao2013

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fuzimiao2013

#20 h Jan 8, 2022, 1:50 AM

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I'm laughing because it's an overkill, but I honestly would LOVE to see more solutions like this - all the intermediate steps, the thought processes - you can get a lot more if you're a beginner.

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Math4Life2020

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Math4Life2020

#21 h Jan 8, 2022, 1:55 AM

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i dare someone to do 2021 imo/3 using 2nd-grade level steps

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asdf334

7585 posts

asdf334

#22 h Jan 8, 2022, 2:14 AM

Y by

DeToasty3 wrote:

We start with a miraculous claim which is the pith of this problem:

Claim 1. Bob's number must be $2$ .

Proof. Before we can get our hands dirty, we shall prove a few subclaims:

Subclaim 1.1. Alice cannot have the number $1$ .

Proof. If she did, then Alice would know that Bob's number is greater than hers.

Subclaim 1.2. Alice cannot have the number $40$ .

Proof. If she did, then Alice would know that Bob's number is less than hers.

Subclaim 1.3. $1$ is a possibility for Bob's number.

Proof. If $1$ is Bob's number, then he knows that Alice's number is larger than his.

Subclaim 1.4. $2$ is a possibility for Bob's number.

Proof. If $2$ is Bob's number, then he knows that, because Alice's number cannot be $1$ , Alice's number is larger than his.

Subclaim 1.5. $39$ is a possibility for Bob's number.

Proof. If $39$ is Bob's number, then he knows that, because Alice's number cannot be $40$ , Alice's number is less than his.

Subclaim 1.6. $40$ is a possibility for Bob's number.

Proof. If $40$ is Bob's number, then he knows that Alice's number is less than his.

We have deduced that Bob's number can be either $1$ , $2$ , $39$ , or $40$ . Of these four numbers, only $2$ is prime, hence we have proved the claim. $\square$

Now that we know that Bob's number is $2$ , we may use Alice's final sentence to finish the problem. Multiplying $2$ by $100$ is merely equivalent to computing the expression $\underbrace{2 + 2 + 2 + \cdots + 2 + 2}_{100\text{ times}}$ , from which we produce the number $200$ . Now, we must search for perfect squares. Before we do this, we have to bound the range of possible values of this perfect square. To do this, we introduce another claim:

Claim 2. The perfect square must be contained in the interval $(202,240)$ .

To prove this claim, we introduce two subclaims.

Subclaim 2.1. The perfect square must be strictly greater than $202$ .

Proof. As Alice cannot have the number $1$ , she must have at least the number $2$ . However, since Bob has the number $2$ , Alice must have at least the number $3$ . Thus, the perfect square must be at least $200+3=203$ , hence proven.

Subclaim 2.2. The perfect square must be strictly less than $240$ .

Proof. As Alice cannot have the number $40$ , she must have at most the number $39$ . Thus, the perfect square must be at most $200+39=239$ , hence proven.

Combining these two subclaims, we have that the perfect square must be contained in the interval $(202,240)$ , as desired. $\square$

Now, we must search for perfect squares in the interval $(202,240)$ . Upon squaring $15$ , we find that the number obtained is $225$ , which, by sheer luck, is in the interval $(202,240)$ . However, upon squaring $14$ and $16$ , we obtain the numbers $196$ and $256$ , neither of which are in the interval. It is well-known that $f(x)=x^2$ is strictly increasing as $x$ increases from $0$ . Thus, if $14^2$ and $16^2$ are out of the interval $(202,240)$ from opposite sides of the interval, we may conclude that $15^2=225$ is the only possible perfect square. This means that Alice's number is $225-200=25$ .

Finally, we may take the sum of Alice's and Bob's numbers, which are $25$ and $2$ , respectively, for a final answer of $25+2=\boxed{\textbf{(A) }27}$ , and we are done. $\blacksquare$

Remark. This problem is clearly unsuitable for the AMC 10. Proving even one of these claims by itself proves to be a significant challenge, perhaps too much for tenth graders.

Truly a marvelous solution, the likes of which have not been seen in a century.

This post has been edited 2 times. Last edited by asdf334, Jan 8, 2022, 2:22 AM

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JH_K2IMO

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JH_K2IMO

#24 h Nov 15, 2024, 8:07 AM

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If Alice had drawn either 1 or 40, she would have immediately known who drew the larger number.
Therefore, Alice’s number is not 1 or 40.
Bob, on the other hand, says that he knows who has the larger number, so Bob must have drawn one of the numbers 1, 2, 39, or 40.
Since Bob's number is a prime number, his number must be 2.
Given that 2 x 100 + (Alice's number) must be a perfect square, Alice’s number must be 25.
Therefore, the sum of their numbers is 2 + 25 = 27 .
The answer is A.

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