The Logical Lie Detection - Puzzle

Three Paley brothers and three Thomson brothers operate a company that manufactures lie detectors. Three of these six men always tell the truth, and three always tell lies; neither set of brothers consists exclusively of liars. 

Some recent statements from the six men are recorded below. 

Can you find the six men's full names, and tell which men tell the truth and which tell lies?

1. Alan: "Both my brothers tell lies."

2. Boris: "Both my brothers tell the truth."

3: Chuck: "Alan and Boris are both liars."

4. Dalman: "Chuck and I are brothers."

5. Edwin: "Boris and I are brothers."

6. Finney: "Edwin tells the truth."

7. Finney: "Boris is one of the Paleys."

Click here is the SOLUTION of the puzzle! 

The Logical Lie Detection - Solution

What was the puzzle?

The statement given by all the six persons are - 

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1. Alan: "Both my brothers tell lies."

2. Boris: "Both my brothers tell the truth."

3: Chuck: "Alan and Boris are both liars."

4. Dalman: "Chuck and I are brothers."

5. Edwin: "Boris and I are brothers."

6. Finney: "Edwin tells the truth."

7. Finney: "Boris is one of the Paleys."

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1] Since as per given information, neither set of brothers consists of entirely all liars (hence truth tellers), 3 liars (or 3 truth tellers) must be distributed as (2, 1 or 1, 2) among 2 groups of 3 brothers.

2] Boris says his both brothers are truth tellers. If his statement is true then there will be 3 brothers telling the truth which is impossible. Hence, Boris is a liar.

3] If Alan's statement (1) is truth then Chuck must be lying in his statement (3). And if Alan is lying then Chuck must be telling the truth. 

 That is one of the Alan or Chuck is a truth teller and other is a liar.

4] So far we got 2 liars (Boris and Alan/Chuck) and 1 truth teller (Chuck/Alan). Since, there are total of 3 liars & 3 truth tellers in total, there must be 2 truth tellers and 1 liar among Dalman, Edwin and Finny.

5] If Finney is lying then his statement (6) suggests that Edward is also liar. But we can't have 2 liars among Dalman, Edwin and Finney as deduced above. Hence, Finny must be telling the truth and hence Edwin.

6] The true statement (5) of Edwin suggests that Edwin and Boris are brothers and as per truth teller Finney, surname of Boris & hence of Edwin is Paley.

7] As Finney is telling the truth (and hence Edwin), the Dalman must be lying. This way, we have 2 truth tellers and 1 liar among Finney, Chuck and Dalman as deduced in step 4.

8] The lying statement (4) of Dalman suggests that Chuck and Dalman are not the brothers. Hence, one of them is Paley and other is Thomson.

9] So third brother of Boris and Edwin must be either Chuck or Dalman. So, Alan and Finney must be brothers having surname Thomson.

10] Since, Finny is telling the truth the statement (1) of Alan (suggesting both of his brothers are liars) must be a lie. 

11] And if Alan is lying then Chuck must be telling the truth (STEP 3).

12] Now, if Chuck is third brother of Boris and Edwin Paley, then statement of Boris (2) would be true and all Paley brothers would be truth tellers which is impossible.

13] Hence, Dalman who is liar (STEP 7) must be third brother of Boris and Edwin Paley.

14] Obviously, since Chuck isn't brother of Dalman, he must have surname Thomson like Alan and Finney.

CONCLUSION : 

Full Name : Alan Thompson,    Behavior : Liar
Full Name : Boris Paley,          Behavior : Liar
Full Name : Chuck Thomson,   Behavior : Truth teller
Full Name : Dalman Paley,      Behavior : Liar
Full Name : Edwin Paley,         Behavior : Truth teller
Full Name : Finney Thomson,   Behavior : Truth teller
 

The First Case of Mystery Number

There is a ten-digit mystery number (no leading 0), represented by ABCDEFGHIJ, where each numeral, 0 through 9, is used once. 

Given the following clues, what is the number?

1) A + B + C + D + E is a multiple of 6.

2) F + G + H + I + J is a multiple of 5.

3) A + C + E + G + I is a multiple of 9.

4) B + D + F + H + J is a multiple of 2.

5) AB is a multiple of 3.

6) CD is a multiple of 4.

7) EF is a multiple of 7.

8) GH is a multiple of 8.

9) IJ is a multiple of 10.

10) FE, HC, and JA are all prime numbers.

NOTE : AB, CD, EF, GH and IJ are the numbers having 2 digits and not product of 2 digits like A and B, C and D .....

HERE is that MYSTERY number! 

Demystifying The First Mystery Number

What was the challenge?

Take a look at the clues given for identifying the number ABCDEFGHIJ.

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1) A + B + C + D + E is a multiple of 6.
 
2) F + G + H + I + J is a multiple of 5.
 
3) A + C + E + G + I is a multiple of 9.
 
4) B + D + F + H + J is a multiple of 2.
 
5) AB is a multiple of 3.
 
6) CD is a multiple of 4.
 
7) EF is a multiple of 7.
 
8) GH is a multiple of 8.
 
9) IJ is a multiple of 10.
 
10) FE, HC, and JA are all prime numbers.

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STEPS :  

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STEP 1 : Since, the digits in number ABCDEFGHIJ are from 0 to 9 with no repeat, the sum of all digits must be 0 + 1 + .....+ 9 = 45.

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STEP 2 : In first 2 conditions, it's clear that all digits of mystery number are added i.e. from A to J. However, addition of first 5 digits is multiple of 6 and addition of rest of digits is multiple of 5. 

That means the total addition of 45 must be divided into 2 parts such that one is multiple of 6 & other is multiple of 5.

30 and 45 is only pair that can satisfy these conditions. Hence,

A + B + C + D + E = 30.

F + G + H + I + J = 15.

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STEP 3 : In next 2 conditions, sums of digits at odd positions and even positions are listed. Moreover, the sum of digits at odd positions has to be multiple of 9 & that of at even positions need to be multiple of 2.

So again,  the total addition of 45 must be divided into 2 parts such that one is multiple of 9 & other is multiple of 2.

The only pair to get these conditions true is 27 and 18. Hence, 

A + C + E + G + I = 27.

B + D + F + H + J = 18.

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STEP 4 : As per condition 9, IJ is multiple of 10. For that, J has to be 0 and with that now 0 can't be anywhere else. J = 0. 

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STEP 5 : Since, one digit can be used only once, numbers like 11, 22, 33....are eliminated straightaway.

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STEP 6 : As per condition 10, JA is prime number. With J = 0, for JA to be prime number, A = 2, 3, 5, 7. 
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STEP 7 : As per condition 5, AB is a multiple of 3. 

Let's list out possible value of AB without any 0, possible digits of A = 2, 3, 5, 7 and excluding numbers having 2 same digits as -

  21, 24, 27, 36, 39, 51, 54, 57, 72, 75, 78. 

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STEP 8 : For numbers FE and HC to be prime (as per condition 10), C and E can't be 0, 5 or even.

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STEP 9 : As per condition 6, CD is multiple of 4 and as per condition 8, GH is multiple of 8. So, D and H has to be even digits.

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STEP 10 : As per condition 6, CD is a multiple of 4. So the possible values of CD without 0, with C not equal to 5 and with odd C, even D -

  12, 16, 32, 36, 72, 76, 92, 96. 

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STEP 11 : As deduced in STEP 3 , B + D + F + H + J = 18.  

With J = 0 and D, H as even digits (STEP 9), both B and F has to odd or even to get to the even total of 18. 

If both of them are even then the total of 

B + D + F + H + J  = 2 + 4 + 6 + 8 + 0 = 20.

which is against our deduction.

Hence, B and F must be odd numbers. 

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STEP 12 : So, possible values of AB deduced in STEP 7 are revised with odd B as -

  21, 27, 39, 51, 57, 75.

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STEP 13 : As per condition 7, EF is a multiple of 7. With F as odd (STEP 11), along with E as odd, not equal to 5 (STEP - 8), possible value of EF are - 

  21, 49, 63, 91.

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STEP 14 : But as per condition 10, FE is PRIME number. Hence, the only possible value of EF from above step is 91. SO, E = 9 and F = 1.

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STEP 15 : Now after 1 and 9 already taken by F and E, possible value of AB in STEP 12 are again revised as - 27, 57, 75. And it's clear that either A or B takes digit 7. So 7 can't be used further.

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STEP 16 : So after 7 taken by A or B, E = 9, F = 1 possible values of CD deduced in STEP 10 are revised as - 32, 36.  Hence, C = 3.

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STEP 17 : With AB = 27, CD can't be 32. And if AB = 27, CD = 36 then,

A + B + C + D + E = 2 + 7 + 3 + 6 + D + 1 = 30.

D = 13.

This value of D is impossible.

Moreover, if CD = 32 and AB = 75 or 57, 

A + B + C + D + E = 5 + 7 + 3 + 2 + D + 1 = 30.

D = 12.

Again, this value of D is invalid. 

Hence, CD = 36 i.e. C = 3 and D = 6 and AB = 57 or 75 but not 27.

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STEP 18 : With AB = 57 or 75, CD = 36, EF = 91, J = 0, possible values of GH which is multiple of 8 (condition 8) are -  24, 48. 

That means either G or H takes 4. Or G is either 2 or 4.

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STEP 19 :  Now as deduced in STEP 3,

A + C + E + G + I = 27. 

A + 3 + 9 + G + I = 27

A + G + I = 15.
 
The letter G must be either 2 or 4 and A may be 5 or 7.

If A = 5, G = 4 then I = 6

If A = 7, G = 2 then I = 6

But we have D = 6 already, hence both of above are invalid.

If A = 7, G = 4 then I = 4.

Again, this is invalid as 2 letters G and I taking same digit 4.

Hence, A = 5, G = 2 is only valid combination thereby giving I = 8.

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STEP 20 : 

If A = 5, then B = 7 ( STEP 17 ). 

C = 3, D = 6 ( STEP 17 ).

E = 9, F = 1 ( STEP 14).

If G = 2, then H = 4 ( STEP 18 & 19).

I = 8 (STEP 19), J = 0 ( STEP 4). 

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CONCLUSION :

Hence, the mystery number ABCDEFGHI is 5736912480.

In the end, just to verify if the number that we have deduced is following all given conditions, 

1) 5 + 7 + 3 + 6 + 9 = 30 is  a multiple of 6.
2) 1 + 2 + 4 + 8 + 0 = 15 is a multiple of 5.
3) 5 + 3 + 9 + 2 + 8 = 27 is a multiple of 9.
4) 7 + 6 + 1 + 4 + 0 = 18 is a multiple of 2.
5) 57 is a multiple of 3.
6) 36 is a multiple of 4.
7) 91 is a multiple of 7.
8) 24 is a multiple of 8.
9) 80 is a multiple of 10.
10) 19, 43, and 05 are prime numbers.

"Which one of the golfers is Mr. Blue?"

Four golfers named Mr. Black, Mr. White, Mr. Brown and Mr. Blue were competing in a tournament. 

The caddy didn't know their names, so he asked them. One of them, Mr. Brown, told a lie.


The 1st golfer said "The 2nd Golfer is Mr. Black."

The 2nd golfer said "I am not Mr. Blue!"

The 3rd golfer said "Mr. White? That's the 4th golfer."

And the 4th golfer remained silent.

Which one of the golfers is Mr. Blue?

Know here who is named as Mr. Blue! 

The Golfer Whose Name is Mr.Blue!

What was the puzzle?

We know that Mr. Brown told a lie and statements of 3 golfers are - 

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The 1st golfer said "The 2nd Golfer is Mr. Black."

The 2nd golfer said "I am not Mr. Blue!"

The 3rd golfer said "Mr. White? That's the 4th golfer."

And the 4th golfer remained silent. 

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Let's name golfers as GOLFER 1, GOLFER 2, GOLFER 3 and GOLFER 4.

1. If we assume the GOLFER 1 is Mr. Brown then his statement must be lie and other 3 must be telling the truth. That is GOLFER 2 must not be Mr. Black and neither Mr. Blue while GOLFER 4 must be Mr. White. 

So, the only name left for GOLFER 2 is Mr. Brown which is already 'occupied' by GOLFER 1 as per our assumption. 

Hence, GOLFER 1 can't be Mr. Brown.

2. Let's suppose the GOLFER 2 himself is Mr. Brown who statement has to be lie. But in his statement he is telling the truth that he is not Mr. Blue. That's contradictory to the given fact that Mr. Brown told a lie.

Hence, GOLFER 2 must not be Mr. Brown.

3. Only golfer left now for the name Mr. Brown is GOLFER 3 who must be lying in his statement. So, the GOLFER 4 must not be Mr. White.

The GOLFER 2 must be Mr. Black as pointed be truly by GOLFER 1 and 'assisted' by true statement made by GOLFER 2.

If GOLFER 2 is Mr. Black, GOLFER 3 is Mr. Brown and GOLFER 4 is not Mr. White then GOLFER 1 must be Mr. White and GOLFER 4 must Mr. Blue.

So the golfer who is named as Mr. Blue is GOLFER 4 i.e. 4th golfer.