Puzzle : Which one is the car thief?

A car thief, who had managed to evade the authorities in the past, unknowingly took the automobile that belonged to Inspector Detweiler. 

The sleuth wasted no time and spared no effort in discovering and carefully examining the available clues. He was able to identify four suspects with certainty that one of them was the culprit.

The four make the statements below. In total, six statements are true and six false.

Suspect A:

1. C and I have met many times before today.
2. B is guilty.
3. The car thief did not know it was the Inspector's car.

Suspect B:

1. D did not do it.
2. D's third statement is false.
3. I am innocent.

Suspect C:
 
1. I have never met A before today.
2. B is not guilty.
3. D knows how to drive.

Suspect D:
 
1. B's first statement is false.
2. I do not know how to drive.
3. A did it.

Which one is the car thief?

Know here who is that car thief? 

Solution : The Unlucky Car Thief

What was the puzzle?

Take a look at the statements made by suspects.

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Suspect A:

1. C and I have met many times before today.
2. B is guilty.
3. The car thief did not know it was the Inspector's car.

Suspect B:

1. D did not do it.
2. D's third statement is false.
3. I am innocent.

Suspect C:

1. I have never met A before today.
2. B is not guilty.
3. D knows how to drive.

Suspect D:

1. B's first statement is false.
2. I do not know how to drive.
3. A did it.

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After investigation, it is found that,  in total, six statements are true and six false.

We will name statements as A1 for first statement of A, A2 for his second statement, B1 for B's first statement B2 for his second statement and so on.

1. Since, a car thief, who had managed to evade the authorities in the past, unknowingly took the automobile that belonged to Inspector Detweiler, we assume statement A3 is TRUE.

2. A1 and C1, C3 and D2 are contradicting statements. These statements are having least relevant in the process as they are not pointing to anybody else. Two of them must be TRUE and 2 must be FALSE. There are 4 TRUE and 4 FALSE statements from rest of statements.

We have, 2 FALSE statements among A1, C1, C3 and D2 for sure.

3. Assume A is a car thief. Then only A2, B2 and D3 turns out to be FALSE from rest giving in total of 5 FALSE statements only.

4. Assume C is a car thief. Then only A2, D1 and D3 are FALSE, hence total of 5 FALSE statement among all statements.

5. Assume D is a car thief. Then again only A2, B1 and D3 are FALSE, once again total 5 out of 12 given statements are FALSE in the case.

6. Assume B is a car thief. In this case, B3, C2, D1 and D3 turns out to be FALSE. Hence, total 6 out of 12 given statements are FALSE. 

This is exactly as per fact found in the investigation which suggests that exactly 6 out of 12 statements are FALSE. 

Hence, B must be a car thief. 

The Case of Fourth 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) B + C + F + J = E + G + H + I = AD
2) B - H = J - G = 3
3) C - F = E - I = 5
4) B * I = AJ

Here are steps demystifying the mystery number! 

The First Case of Mystery Number 

The Second Case of Mystery Number 

The Third Case of Mystery Number

Demystefying The Fourth Mystery Number

What was the challenge? 

Given are hints to identify number ABCDEFGHIJ.

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1) B + C + F + J = E + G + H + I = AD

2) B - H = J - G = 3

3) C - F = E - I = 5

4) B * I = AJ

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STEPS : 

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STEP 1 :

The sum of digits from 0 to 9 is 45.

Maximum value of AD = 98 and 

Minimum value of AD = 10 (A can't be 0 as leading 0's not allowed).

If AD = 98 then sum of rest of digits B + C + F + J + E + G + H + I must be 
45 - (9 + 8) = 28.

If AD = 10 then sum of rest of digits B + C + F + J + E + G + H + I must be 
45 - (1 + 0) = 44.

The sum of such 8 digits is divided into 2 parts in form of
B + C + F + J  and E + G + H + I which in turn must be equal to AD.

Therefore, each group of four digits must sum to one of the following: 14, 15, 16, 17, 18, 19, 20, 21, 22 (with AD varying from 28 to 44).

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STEP 2 :

Using Trial And Error method to get possible values of AD.

If AD = 14 then B + C + F + J + E + G + H + I = 45 - (1 + 4) = 40 and

B + C + F + J  = E + G + H + I = 40/2 = 20 = AD but AD = 14 assumed.

Hence, this value of AD is invalid.

Similarly, 15, 16, 17, 19, 20, 22 are invalid values of AD leaving behind only 18 and 21 as possible values.

Hence, A must be either 1 or 2 and D must be either 1 or 8.

That is either A or D takes 1.

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STEP 3 : 

As per Hint 2, B and J > H and G by 3 respectively and since 1 already taken by A or D,

Possible Values of B and J - 3, 5, 6, 7, 8, 9.

Possible Values of H and G - 2, 3, 4, 5, 6.

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STEP 4 : 

As per Hint 3, C and E > F and I by 3 respectively and since 1 already taken by A or D,

Possible Values of C and E - 5, 7, 8, 9.

Possible Values of F and I -  0, 2, 3, 4.

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STEP 5 : 

So with B having possible values as 3, 5, 6, 7, 8, 9 and I having possible values as 0, 2, 3, 4 the equation B * I has following 24 possibilities - 

(3 x 0 = 0) (3 x 2 = 6) (3 x 3 = 9) (3 x 4 = 12)

(5 x 0 = 0) (5 x 2 = 10) (5 x 3 = 15) (5 x 4 = 20) 
  
(6 x 0 = 0) (6 x 2 = 12) (6 x 3 = 18) (6 x 4 = 24)

(7 x 0 = 0) (7 x 2 = 14) (7 x 3 = 21) (7 x 4 = 28) 

(8 x 0 = 0) (8 x 2 = 16) (8 x 3 = 24) (8 x 4 = 32) 

(9 x 0 = 0) (9 x 2 = 18) (9 x 3 = 27) (9 x 4 = 36).

B * I can't be 0 as AJ can't be 0. Also, since A can't be 0 the product B*I can't be single digit like 6 or 9. 

Moreover, A has to be either 1 or 2 as deduced in STEP 2 and J must be among 3, 5, 6, 7, 8, 9 as deduced in STEP 3. 

Now the equation B * I = AJ has possibilities as - 

(5 x 3 = 15) (6 x 3 = 18)
  
(7 x 4 = 28) (8 x 2 = 16) 

(9 x 2 = 18) (9 x 3 = 27)  

Since (5 x 3 = 15) suggests that B = J = 5 which is against the rule that no 2 alphabets can take same digit. Hence, that possibility is eliminated.

Revised Possible Values of B - 6, 7, 8, 9.

Revised Possible Values of I - 2, 3, 4.

Revised Possible Values of J - 6, 7, 8. 

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STEP 6 :  

Since I can't be 0, E can't be 5 as E - I = 5.

Revised Possible Values of E - 7, 8, 9.

Since J - G = 3, and if J is among 6, 7, 8

Revised Possible Values of G - 3, 4, 5.

Since B - H = 3, and if B is among 6, 7, 8, 9

Revised Possible Values of H - 3, 4, 5, 6.

So letters A, B, D, E, G, H, I and J together takes digits 1, 2, 3, 4, 6, 7, 8 and 9 not in order though.

This leaves behind only possible value of C = 5 and F = 0. 

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STEP 7 :

Now, since H can't be 5 hence B can't be 8. Also, G too can't be 5 so J can't be 8 too. So both B and J can't be 8. 

Now, revising B * I = AJ possibilities deduced in STEP 4 as - 

 (9 x 3 = 27)  

Leaves only possible valid combination thereby.

So, we get, B = 9, I = 3, A = 2 and J = 7.

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STEP 8 : 

If B = 9 then H = 6.

If I = 3 then E = 8.

If J = 7 then G = 4.

The equation B + C + F + J = 9 + 5 + 0 + 7 = 21 = AD gives A = 2 and D = 1.

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CONCLUSION :

A = 2, B = 9, C = 5, D = 1, E = 8, F = 0, G = 4, H = 6, I = 3, J = 7.

Hence, the mystery number ABCDEFGHIJ is 2951804637.

 Verifying the given hints - 

1) B + C + F + J = E + G + H + I = AD     
    9 + 5 + 0 + 7 = 8 + 4 + 6 + 3 = 21

2) B - H = J - G = 3
    8 - 3 = 5 - 0 = 5 

3) C - F = E - I = 5
    7 - 4 = 9 - 6 = 3 

4) B * I = AJ 
    9 * 3 = 27