BrainDen.com - Brain Teasers

## Question

Each of eight logicians has a unique positive integer less than 100 stapled to their foreheads. They stand in a circle facing one another so every logician can see everyone's number, save for their own. Each logician must discern their number and raise their hand. After all logicians have raised a hand, they must then all declare their number in unison. What strategy (within the assumed spirit of these types of puzzles) might they employ such that all are correct?

EDIT

## Recommended Posts

• 0

Am guessing that they can agree on a system of communication beforehand

Each logician communicates with the other directly opposite by raising his hand in a precise way. Fingers represent binary values. Left hand, right hand, palm out or in represent four meanings. Perhaps add from 0, add from 33, add from 66 & subtract from 100. In this way each is told his number. I'll keep my fingers crossed.

I like the image of the "stapled to head"

##### Share on other sites

• 0

I thought for sure I mentioned in the OP that they were attending the Arctic Circle of Logicians annual Fresh Air/Fresh Ideas Seminar and were all wearing mittens. Apparantly not, so your answer does indeed fit within the constraints of the OP, Time Out. But for grins, let's do assume the logicians are all wearing plain white mittens.

##### Share on other sites

• 0

I thought for sure I mentioned in the OP that they were attending the Arctic Circle of Logicians annual Fresh Air/Fresh Ideas Seminar and were all wearing mittens. Apparantly not, so your answer does indeed fit within the constraints of the OP, Time Out. But for grins, let's do assume the logicians are all wearing plain white mittens.

When you say 'numbers less than 100', do you mean 'integers less than 100 but larger than 0'? Also, can a number be repeated?

Edited by bushindo
##### Share on other sites

• 0

(assumming numbers cannot be repeated) each logician will look at the person with the smallest number on his head, the person with the smallest number on his head will have 7 people looking at him, so he will know that his number is smaller than the rest, that's all I could think of now.

or I could go for the simpler way

(assuming all numbers are integers) blinking system, each person will blink the number of the person standing directly in front of him, to make things easier make 2 sets of blinks, one for tens and on for units, zeros can be winks, if no winking is allowed, then go for normal blinking, though it's not too practical to blink 100 times

##### Share on other sites

• 0

hey bushindo - right you are, rewrote this this a.m. and forgot that bit. hadn't saved my .txt draft and Microsoft was kind enough to update and reboot my machine last night. thanks and sorry

and sorry, too - no blinking in morse code, looking in the reflection of the irises to your right, hoof stomping, shouting the number across from you in pig latin, etc., etc...

##### Share on other sites

• 0

Once a hand is raised, can it be put down? Or would that qualify as illegal communication?

##### Share on other sites

• 0

hey Mo Ma - no, once a hand is raised that is the only bit of info the rest of the group can directly discern from that logician.

##### Share on other sites

• 0

Each of eight logicians has a unique positive integer less than 100 stapled to their foreheads. They stand in a circle facing one another so every logician can see everyone's number, save for their own. Each logician must discern their number and raise their hand. After all logicians have raised a hand, they must then all declare their number in unison. What strategy (within the assumed spirit of these types of puzzles) might they employ such that all are correct?

EDIT

Here's an attempt.

Let the logicians be indexed by numbers going from 1 to 8. Let aj represent the positive integer of logician j. Let di( b ) represent the i-th digit from the right in the binary expansion of b. For instance, if b = 99 (which is 1100011 in binary), then d2( b) = 1, d3( b ) = 0, d4(b) = 0, and so on.

The general strategy is as follows.

1) For each logician i, where 1<= i <=7, look around and sum up the remaining hats. Let's call this sum Ti. The logician i is then supposed to raise his hand if di( Ti) = 1. Otherwise the logician should not raise his hand.

2) After logician 1 to 7 have done the previous procedure, then logician 8 should be able to figure out his number.

3) Logician 8 then calculates the sum ( d1( a1) + d2( a2) + ... + d7(a7) ). He then should raise his hand if this sum is odd. The remaining 7 logicians should be able to figure out his number as well.

Edited by bushindo
• 1
##### Share on other sites

• 0

I just want to come back and say that this is a really well designed puzzle. I didn't explain the rationale behind the solution before, and so here it is

The rationale of the solution is that we want to use the logicians 1-7 to convey (in binary digits) to the 8th logician his number. We need 7 binary digits to describe a number between 1 and 100, hence the 7 logicians.

The procedure that we did above will also allow each of logician 1 to 7 to determine six of his seven binary digits. We then simply used the 8th logician to convey this information back to the previous 6.

Edited by bushindo
##### Share on other sites

• 0

if we assume they can count inside at same rhythm problem would be too easy so i assume not

##### Share on other sites

• 0

Here's an attempt.

Let the logicians be indexed by numbers going from 1 to 8. Let aj represent the positive integer of logician j. Let di( b ) represent the i-th digit from the right in the binary expansion of b. For instance, if b = 99 (which is 1100011 in binary), then d2( b) = 1, d3( b ) = 0, d4(b) = 0, and so on.

The general strategy is as follows.

1) For each logician i, where 1<= i <=7, look around and sum up the remaining hats. Let's call this sum Ti. The logician i is then supposed to raise his hand if di( Ti) = 1. Otherwise the logician should not raise his hand.

2) After logician 1 to 7 have done the previous procedure, then logician 8 should be able to figure out his number.

3) Logician 8 then calculates the sum ( d1( a1) + d2( a2) + ... + d7(a7) ). He then should raise his hand if this sum is odd. The remaining 7 logicians should be able to figure out his number as well.

yes indeed, bushindo. well done.

all 8 logicians could perform your step 3) above. Then every digit of each logician will appear exactly once on each of the seven other logicians step 3). And each logician can already see six of those seven digits.

for example:

logician(1) step 3)-->d8(a8)+d7(a7)+…+d2(a2) and logician(8) can see d7(a7)+d6(a6)+…+d2(a2) so knowing the parity of logician(1)'s step 3), logician(8) can easily figure her eighth digit, etc.

##### Share on other sites

• 0

yes indeed, bushindo. well done.

all 8 logicians could perform your step 3) above. Then every digit of each logician will appear exactly once on each of the seven other logicians step 3). And each logician can already see six of those seven digits.

for example:

logician(1) step 3)-->d8(a8)+d7(a7)+…+d2(a2) and logician(8) can see d7(a7)+d6(a6)+…+d2(a2) so knowing the parity of logician(1)'s step 3), logician(8) can easily figure her eighth digit, etc.

That alternative solution is a lot more compact and elegant. Thanks for sharing this problem.

##### Share on other sites

• 0

hmm, not convinced its as simple as bushindo/plain makes it out to be.

let's say everyone has an even binary parity.

1 2 3 4 5 6 7 8

3 6 12 15 24 30 48 60

0000011

0000110

0001100

0001111

0011000

0011110

0110000

0111100

does no one raise their hand? or even if some poeple raise, thare are various ways to get all even binary parities. how would everyone know thier number?

##### Share on other sites

• 0

The way I look at it, the ith person raises her hand if and only if

d1(a1+(i mod 8)) ⊕ d2(a1+(i+1 mod 8)) ⊕ d3(a1+(i+2 mod 8)) ⊕ d4(a1+(i+3 mod 8))

⊕ d5(a1+(i+4 mod 8)) ⊕ d6(a1+(i+5 mod 8)) ⊕ d7(a1+(i+6 mod 8)) = 1

(where ⊕ is addition modulo 2). Using knowledge of the numbers on the heads

you can see, as well as which hands are raised, each person can determine her

number.

##### Share on other sites

• 0

yes sp, thanks for the explanation. and where hi is the sum mod(2) determining if logician i raises a hand or not (or raises a left or right mittened paw)

so to figure say a8 -

to find the seventh digit logician 8 can see six digits of h1 (colored buff) whose sum mod(2)=0=h1 so the seventh digit of a8=0

to find the sixth digit logician 8 can see six digits of h2 (colored yellow) whose sum mod(2)=1 but h2=0 so the sixth digit of a8=1

etc.

I think the solution still works if the word "unique" is removed from the problem statement. N'est pas?

et @ superprismatique - mais bien sur

zero works as well as integers up to 127 but thought limiting to unique two digit integers might better obscure the path to the solution • 1
##### Share on other sites

• 0

An alternative solution:

Before being placed in a circle, all agree on a march to hum inside their heads, and synchronize their humming. Then they go silent when placed in a circle, but keep humming and counting lines in their heads. This is MUCH more accurate than counting in words.

Each person raises a hand on the line # of the person opposite them.

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account. ×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

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