Jump to content
BrainDen.com - Brain Teasers
  • 0

Guessing right number from 1 to 27, with 3 Yes/No/Maybe questions.


jasen
 Share

Question

Your best friend (Bob) comes to you with this problem.

The Problem

His teacher (Mr. Sam) will ask Bob to work out a number in a child (Joan) head from 1 and 27.
 Bob has to find the correct number, by asking Joan three Yes/No questions about the number.
 Joan can respond with 3 different answers: "Yes", "No", or "I do not know".

Joan is a smart boy, but he doesn't know about another base number except 10. So by asking that kind of question he will answer "I do not know"

Example:
If the number is 12

Is it an even number? Yes.
 Is it divisible by 9? No.
 If I take a random number between 10 to 15, will I get the right number? I do not know.

------------------

Another Example:
If the number is 9


Is it an even number? No.
 Is it divisible by 9? Yes.
 If I take a random number between 10 to 15, will I get the right number? No.

------------------
 

Find the strategy, so every number from 1 to 27 can be guessed correctly.
 

Link to comment
Share on other sites

3 answers to this question

Recommended Posts

  • 0
Spoiler

Let's say the correct number is x

Divide the numbers into three equal-size groups

A: { 1  2  3  4  5  6  7  8  9}
B: {10 11 12 13 14 15 16 17 18}
C: {19 20 21 22 23 24 25 26 27}

Ask Q1: If I pick a number at random from B, (the middle group), will it be greater than x?
Answers (Yes / IDK / No) indicate x is in (A/B/C) respectively.
Suppose Joan answered Yes. Then
x is in A. Divide A into three equal-size groups:

D: {1 2 3}
E: {4 5 6}
F: {7 8 9)

Ask Q2: if I pick a number at random from E, (the middle group) will it be greater than x?
Answers (Yes / IDK / No) indicate x is in (D/E/F) respectively.
Suppose Joan answers IDK. Then
x is in E, and therefore is 4, 5, or 6.

Now we need a question different in form from the first two. In order for IDK to be a possible response, randomness has to be present. Since there is no randomness in a number picked at random from a single-element set, we need to get creative.

Ask Q3: if I were to pick a number at random from E , (the group identified by Q2),that turned out not to be x, would it be greater than x?
Answers (Yes / IDK / No) indicate x is (4 / 5 / 6) respectively.

So we can identify the correct number using three questions. The first two ask about a number picked at random from the  middle of three multiple-number groups. The third asks about a number picked at random, from the group of three numbers that survives Q2, that turns out not to be x..

Summary:

  1. Divide the numbers into three groups, as above. Ask: If I were to select a number at random from the middle group, would it be greater than x? Select the (first / second / third) group if the answer is (Yes / IDK / No).
     
  2. Divide the selected group into three groups, as above. Ask (again): If I were to select at random a number from the middle group, would it be greater than x? Select the (first / second / third) group if the answer is (Yes / IDK / No).
     
  3. Finally, ask: If I were to select at random a number from (the last selected group) that turned out not to be x, would it be greater than x? Select the (first / second / third) number from the group if the answer is (Yes / IDK / No).

 

Link to comment
Share on other sites

  • 0

First thoughts:

Spoiler

It sounds like the weighing problem where one of 27 coins is counterfeit, differing in weight from all the others.

The difference is that a set of balances can give 3 different outcomes, whereas a normal yes/no question gives only two. But IDK has been added to the mix. So perhaps the solution hinges on finding a way to use {yes, no, IDK} responses in an aalogous way to using {left-heavier, equal, right-heavier} for the weighing problem.

If with a single question we can deduce which of {1 2 3} was chosen, then we should with three questions be able to determine which of {1 2 3 4 ,,, 25 26 27} was chosen.

 

Link to comment
Share on other sites

  • 0

 

Spoiler

Here is my version :

I  will create 3 pairs of list

First pair : 
1A : 2,3,5,6,8,9,11,12,14,15,17,18,20,21,23,24,26,27
1B : 2,5,8,11,14,17,20,23,26

Then ask Mr. Smith : If I take a random list, Does the number in the list ?
If the answer is Yes then, write number 2 in first paper.
(because the number appears in both list)
If the answer is No then, write number 0 in first paper.
(because the number do not appears in both list)
If the answer is I do not know then, write number 1 in first paper.
(because the number only appears in one list)

then

2nd pair :
2A : 4,5,6,7,8,9,13,14,15,16,17,18,22,23,24,25,26,27
2B : 4,5,6,13,14,15,23,24,25

Then ask Mr. Smith : If I take a random list, Does the number in the list ?
If the answer is Yes then, write number 2 in the 2nd paper.
If the answer is No then, write number 0 in the 2nd paper.
If the answer is I do not know then, write number 1 in the 2nd paper.

then

3rd pair :
3A : 10 to 27
3B : 10 to 18 

Then ask Mr. Smith : If I take a random list, Does the number in the list ?
If the answer is Yes then, write number 2 in the 3rd paper.
If the answer is No then, write number 0 in the 3rd paper.
If the answer is I do not know then, write number 1 in the 3rd paper.

then

put the number in the order 3rd 2nd 1st as base 3 number, than convert it into base 10 number,do not forget to add 1. The result is Mr. Smith number.

 

 

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...