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.

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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. 

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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.

 

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 - 

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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.


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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.