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.

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

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.

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

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. 

Crack The 10-Digits Password

In an attempt to further protect his secret chicken recipe, Colonel Sanders has locked it away in a safe with a 10-digit password.  Unable to resist his crispy fried chicken, you'd like to crack the code. 

Try your luck using the following information:

All digits from 0 to 9 are used exactly once.

No digit is the same as the position which it occupies in the sequence.

The sum of the 5th and 10th digits is a square number other than 9.

The sum of the 9th and 10th digits is a cube.

The 2nd digit is an even number.

Zero is in the 3rd position.

The sum of the 4th, 6th and 8th digits is a single digit number.

The sum of the 2nd and 5th digits is a triangle number.

Number 8 is 2 spaces away from the number 9 (one in between).

The number 3 is next to the 9 but not the zero.

Click here to know how to CRACK it! 

Cracking The 10-Digits Password

What was the challenge? 

Let's suppose the 10-digits password is ABCDEFGHIJ.

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

Take a look at the given clues to crack down the password.

1. All digits from 0 to 9 are used exactly once.

2. No digit is the same as the position which it occupies in the sequence.

3. The sum of the 5th and 10th digits is a square number other than 9.
 
4. The sum of the 9th and 10th digits is a cube.
 
5. The 2nd digit is an even number.
 
6. Zero is in the 3rd position.
 
7. The sum of the 4th, 6th and 8th digits is a single digit number.
 
8. The sum of the 2nd and 5th digits is a triangle number.
 
9. Number 8 is 2 spaces away from the number 9 (one in between).
 
10. The number 3 is next to the 9 but not the zero. 

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

STEPS : 

Respective Clues are in ().

1] Since, every digit is used only once (1), the sum of 2 digits can't exceed 
9 + 8 = 17 and minimum sum of 2 digits can be only 0 + 1 = 1.

2] So, the sum of 5th & 10th digits which is square (3) can be - 1, 4, 16.
The sum of 9th & 10th digits which is cube (4) can be - 1, 8.
The sum of 2nd & 5th digits which is triangle number (8) can be - 1, 3, 6, 10, 15.

3] As per (6), 0 is already at third position i.e. C = 0. Hence, 2nd, 5th, 9th or 10th digits can't be 0. Therefore, the sum of 5th & 10th or 9th & 10th or 2nd & 5th can't be 1. Revising list of possible values of those in step 2.

The sum of 5th & 10th digits which is square (3) can be -  4, 16.
The sum of 9th & 10th digits which is cube (4) can be -  8.
The sum of 2nd & 5th digits which is triangle number (8) can be -  3, 6, 10, 15.

4] Possible pairs of 9th & 10th digits - (1,7), (7,1), (2,6), (6,2), (3,5), (5,3)

If IJ = 71, then 5th digit E must be 4 - 1 = 3 & possible values of 2nd digit B are 0,3,7.
But as per (5), B must be even & C = 0 already.

IJ = 26 or 62 or 35 doesn't leave any valid value for 5th digit E.

If IJ = 53, then 5th digit E must be 1 & possible values of 2nd digit B are 2,5,9
That is B = 2 (5). But as per (2), no digit is the same as the position which it occupies in the sequence. So, the number 2 can't be at 2nd position where B is there.

Hence, IJ must be 17 i.e. I = 1, J = 7.

5] This leaves only 9 as valid value for 5th digit E so that (3) is true. 

So, E = 9 and hence B = 6 with sum of 2nd digit B and 5th digit E as 15.

6] So far, we have, B = 6, C = 0, E = 9, I = 1 and J = 7. 

As per (10), the number 3 is next to the 9 but not the zero. Hence, the 6th digit F must be 3. F = 3.

7] As per (9), the 8 is 2 spaces away from 9 i.e. here 5th digit E=9. With 3rd position already occupied by C = 0, the digit 8 must be at 7th position. 
So, G = 8. 

8] As per (7), D + F + H must be single digit. We have, F = 3 already, 
so possible values of D + H are 0, 1, 2, 3, 4, 5, 6.

D + H can't be 0 (0+0), 1 (1+0), 2(1+1/2+0), 3(1+2/3+0), 4(2+2/3+1/4+0), 5(2+3/4+1/5+0) since digits are repeated with C=0, I=1, F=3 already.

Hence, D + H = 6. 

9] Now, D = 5 and H = 1 is not possible. Also, D/H can't be 0 or 6. Both D and H can't be 3 at the same time. D can't be 4 as D is at 4th position. 

Hence, D = 2, H = 4.

10] With B = 6, C = 0, D = 2, E = 9, F = 3, G = 8, H = 4, I = 1, J = 7, the only digit left for the first position A is 5. So A = 5.

CONCLUSION : 

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

So, the 10-digits password ABCDEFGHIJ is 5602938417.

 

A New Word Every Day!

During a six-day period from Monday through Saturday, Eliza Pseudonym and her friends Anna, Barbra, Carla, Delilah, and Fiona have subscribed to an internet mailing list that features a new word every day. 

No two women subscribed on the same day. On each day during the six-day period, a different word has been featured (abulia, betise, caryatid, dehisce, euhemerism, and floruit, in some order). 

From the clues below, determine the day on which each woman subscribed, and the day on which each word was featured.

1. Exactly one of the women has a name beginning with the same letter of the alphabet as the word featured on the day that she subscribed to the mailing list.

2. The word "caryatid" was featured precisely two days prior to Fiona joining the mailing list.

3. Carla joined the mailing list on Friday.

4. Anna signed up for the mailing list precisely one day after the word "euhemerism" was highlighted.

5. Wednesday's word did not end with the letter "e".

6. Barbra subscribed precisely three days after the word "dehisce" was featured.

Here is every word of woman of the day! 

Assigned Word to the Woman of the Day

What was the challenge?

As we know, Eliza Pseudonym and her friends Anna, Barbra, Carla, Delilah, and Fiona have subscribed to an internet mailing list that features a new word every day. No two women subscribed on the same day. On each day during the six-day period, a different word has been featured (abulia, betise, caryatid, dehisce, euhemerism, and floruit, in some order).

And given clues are - 

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

1. Exactly one of the women has a name beginning with the same letter of the alphabet as the word featured on the day that she subscribed to the mailing list.

2. The word "caryatid" was featured precisely two days prior to Fiona joining the mailing list.

3. Carla joined the mailing list on Friday.

4. Anna signed up for the mailing list precisely one day after the word "euhemerism" was highlighted.

5. Wednesday's word did not end with the letter "e".

6. Barbra subscribed precisely three days after the word "dehisce" was featured.

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

Let's make a table like below and fill it one by one as per clues.

 STEPS :  

1] As per (3), Carla joined the mailing list on Friday. And as per (6), Barbra's subscription day must be Thursday, Friday or Saturday with word 'dehisce' on Monday or Tuesday or Wednesday. 

But as per (5), ''dehisce'' can't be on Wednesday and Carla has already joined on Friday, hence Barbra must have joined on Thurday with word "dehisce" featured on Monday.

2] Only possible location for "caryatid" and Fiona for (2) to be true are Thursday and Saturday respectively. That's because Monday is already 'occupied' by "dehisce" so Fiona can't be on Wednesday. And as we can see, Thursday and Friday already 'occupied' by Barbra and Carla.

3] With that, for (4) to be true, only possible location for "euhemerism" is Tuesday and for Anna is Wednesday.

4] As per (1), Delilah and Eliza can't be on Monday & Tuesday at a time as that will violate (1). Therefore, Delilah must be on Tuesday and Eliza on Monday.

 
5] As per (5), "betise can't appear on Wednesday. 

Suppose it appears on Friday. 

CASE 1 : The word "abulia" with Anna on Wednesday & "floruit" with Fiona on Saturday. This violates (1), as there would be 2 women have names beginning with the same letter of the alphabet as the word featured on the day that she subscribed to the mailing list.

CASE 2 :  The word "floruit" with Anna on Wednesday & "abulia" with Fiona on Saturday. Again, this is against (1), as there is no woman has a name beginning with the same letter of the alphabet as the word featured on the day that she subscribed to the mailing list.

Therefore, "betise" must be appearing on Saturday, "floruit" on Friday and "abulia" on Wednesday with Anna.

CONCLUSION :

The final table looks like -

The Thanksgiving Pageant

Allison, Jerry, Bonnie, and Bill are participating in a Thanksgiving Pageant at their school. 

The students will portray an Indian warrior, a pilgrim, an Indian maiden, and a deer. Their props consist of a pumpkin, a fish, a basket of corn, and colored leaves. During the pageant, they will recite a poem, sing a song, do a dance, and act as the narrator. Their parents, whose lasts names are Lee, Newton, Myers, and Schuler, will be watching in the audience. The ages of the performers are 13, 12, 11, and 10 years old. 

Use the clues to help you find out who will be doing what for the pageant.

1. Mrs. Lee made her daughter's costume and also the costumes of the Indian maiden and the dancer. 

2. Mr. Myers helped his son rehearse his poem. 

3. The dancer was older than Lee and the pilgrim, but younger than Allison. 

4. The 11-year-old ripped her deer outfit, but her mother pinned it.

5. Newton carried his fish during his tribal dance.

6. Jerry's pumpkin was a symbol of the feast for the pilgrims. 

7. The Indian maiden carried corn as she sang the song of feasting.

Simplified Solution Here! 

Little Performers at The Thanksgiving Pageant

What was the puzzle?

Given Data : 

STUDENTS     : Allison, Jerry, Bonnie, Bill

CHARACTERS : Indian warrior, Pilgrim, Indian maiden, Deer

PROPS           : Pumpkin, Fish, Basket of corn, Colored leaves

ACT                : Poem, Song, Dance, Narrator

LAST NAMES  : Lee, Newton, Myers, Schuler

AGE                : 13, 12, 11, 10

Given Clues : 

1. Mrs. Lee made her daughter's costume and also the costumes of the Indian maiden and the dancer. 

2. Mr. Myers helped his son rehearse his poem. 

3. The dancer was older than Lee and the pilgrim, but younger than Allison. 

4. The 11-year-old ripped her deer outfit, but her mother pinned it.

5. Newton carried his fish during his tribal dance.

6. Jerry's pumpkin was a symbol of the feast for the pilgrims. 

7. The Indian maiden carried corn as she sang the song of feasting.

STEPS :

First of all let's make a table like below & fill it as per clues.

1]  As per clue (1), student having last name Lee, an Indian maiden and dancer are three different students.

2] As per (3), the dancer must be 12 years old with Lee and Pilgrim having age 10 & 11 but order yet to be known. With that Allison must be 13 years old. This clue also suggests that the dancer or Lee isn't pilgrim.

3] The clue (5) clearly suggest that Newton is dancer carrying fish as a property. Also, clue (6) suggests Jerry is pilgrim having pumpkin. And clue (7) suggests that, Indian Maiden sung a song while carrying a basket of corn.

4] With that, we can conclude that Lee must be carrying colored leaves. Also, the clue (4) must be pointing at Lee now. So Lee is 11 years old wearing deer outfit.

5] It's clear now that Newton is Indian Warrior. And, 10 years old (STEP 2) Jerry is Myers who father helped him to rehearse his poem as per clue (2).

6] Since Allison is 13 years old, she must be Indian Maiden carrying a basket of corn while singing a song. She must have last name Schuler - the only last name left. Since (1) suggests Lee is girl, her name must be Bonnie who is narrator carrying a colored leaves. Finally, first name of Newton must be Bill.

7] So, the final table looks as - 

Story of Four High School Friends

Four high school friends, one named Cathy, were about to go to college. Their last names were Williams, Burbank, Collins, and Gunderson. Each enrolled in a different college, one of them being a state college. 

From the clues below determine each person's full name, and the college he or she attended.

1. No student's first name begins with the first letter as her or his last name, and no students first name's last letter is the same as his last name's last letter.

2. Neither Hank or Williams went to the community college.

3. Alan, Collins, and the student who went to the university all lived on the same street. The other student lived two blocks away.

4. Gladys and Hank lived next door to each other.

5. The private college accepted Hank's application, but he decided he could not afford to go there.

Here is ANALYSIS of the story! 

Analysing The Story of Four High School Friends

What was the story?

GIVEN DATA : 

First Name : Hank, Gladys, Cathy, Alan 
Last Name : Collins, Burbank, Gunderson, Williams 
College       :  State College, University, Community College, Private College 

HINTS : 

1. No student's first name begins with the first letter as her or his last name, and no students first name's last letter is the same as his last name's last letter.

2. Neither Hank or Williams went to the community college.

3. Alan, Collins, and the student who went to the university all lived on the same street. The other student lived two blocks away.

4. Gladys and Hank lived next door to each other.

5. The private college accepted Hank's application, but he decided he could not afford to go there.   

STEPS :  

1] Let's make a table like below and fill it as per hints.

  
2] As per Hint (3), we have, Alan, Collins and student going to the university as 3 different students.

3] As per (1), Collins can't be Cathy & as per (4), Cathy can't be at no.3 as in that case no blocks will be left for (4) to be true. Hence Cathy must be at no.4.

4] As per (2), Hank and Williams are two different students. And as per (4), Gladys and Hank must be at 2 & 3 (and anyhow these are only blocks left for them) but order yet to be known. So, Williams is certainly not at 3 or 2. And as per (1), Gladys can't be Gunderson or Collins or Williams & Hank can't be Burbank.  Therefore, Hank must be Collins at 2 and Gladys must be Burbank at 3.

5] Now, as per (1), Alan can't be Gunderson hence must be Williams and Cathy must be Gunderson. 

6] As per (2), neither Hank nor Williams went to community college, hence Cathy Gunderson must be. And as per (5), Hank didn't choose private college hence must have chosen state college while Alan Williams must be in private college.

7] Therefore, the final table looks like as below.

 

Crack And Win $500,000 - Puzzle

You are in a game show with four other contestants. The objective is to crack the combination of the safe using the clues, and the first person to do so will win $500,000.

The safe combination looks like this:

??-??-??-??

A digit can be used more than once in the code, and there are no leading zeroes.

Here are the clues:

1. The third set of two numbers is the same as the first set reversed.

2. There are no 2's, 3's, 4's or 5's in any of the combination.

3. If you multiply 4 by the third set of numbers, you will get the fourth set of numbers.

4. If you add the first number in the first set with the first number in the second set you will get 8.

5. The second set of numbers is not greater than 20.

6. The second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set.

And get moving, I think another contestant has almost figured it out!

Click here for the SOLUTION! 

Crack And Win $500,000 Puzzle - Solution

What was the puzzle?

We know, the safe has a lock having 4 sets of 2 digits as -

?? - ?? - ?? - ??

Since, leading zeros are not allowed any of the set can't be started with 0 like 01, 07, 09 etc. 

Take a look at the clues given - 

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

1. The third set of two numbers is the same as the first set reversed.

2. There are no 2's, 3's, 4's or 5's in any of the combination.

3. If you multiply 4 by the third set of numbers, you will get the fourth set of numbers.

4. If you add the first number in the first set with the first number in the second set you will get 8.

5. The second set of numbers is not greater than 20.

6. The second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set.


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

STEPS :

1] As per clue (2), digits 2, 3, 4 or 5 aren't allowed in any set. That means only digits 0, 1, 6, 7, 8, 9 are allowed for sure.

2] For (3) to be true, possible combinations of 3rd and 4th sets are - 

10 x 4 = 40
11 x 4 = 44
16 x 4 = 64
17 x 4 = 68
18 x 4 = 72
19 x 4 = 76

The third set can't be 10 for 2 reasons. - 1} As per clue (1), the first set will be 01 and leading 0's are not allowed. 2} The 4th set will have digit 4 which is not allowed.

The third set can't start with 2X, 3X, 4X or 5X as those digits aren't allowed. Moreover, it can't be started with 6X, 7X, 8X as in that case the value of the 4th set will exceed it's maximum possible value of 96 (if digit 2 was allowed) or 76.

Out of all above combinations, only 17 x 4 = 68 and 19 x 4 = 76 are valid combination as rest of combinations have digits that aren't allowed.

So, one thing is sure that the first digit of the third set is 1. And hence the first digit of first set also must be 1 as per clue (1).

3] As per clue (5), the second set can't exceed 20. It can't start with 0. Hence, possible values ranges from 10 to 19. That means, the first digit of the second set is also 1.

4] Now as per clue (4), if you add the first number in the first set with the first number in the second set you will get 8. That means the first number of first set is 8 - 1 = 7. 

As of now, the code looks like : 71 - 1? - 1? - ??

5] If 7 is the first digit of first set then as per clue (1), 7 itself is second digit of the third set.

Now, the code looks as : 71 - 1? - 17 - ??

6] As per clue (3) and rightly deduced as a possible combinations for 3rd & 4th set in STEP 2, the 4th set must be 17 x 4 = 68.

With that, code turns into : 71 - 1? - 17 - 68.

7] Finally, as per clue (6), the second number in the second set multiplied by the second number in the fourth set equals one higher than the numbers in the first set. This multiplication i.e. ? x 7 must be equal to 71 + 1 = 72. 
Hence, ? = 9.

The final code looks like : 71 - 19 - 17 - 68.

  

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.

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

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

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

STEPS : 

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


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).

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

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.

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

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.

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

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.

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

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. 

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

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. 

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

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.

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

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.

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

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