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.

 

The Third 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) Digit A is either a square number or a triangle number, but not both.

2) Digit B is either an even number or a cube number, but not both.

3) Digit C is either a cube number or a triangle number, but not both.

4) Digit D is either an odd number or a square number, but not both.

5) Digit E is either an odd number or a cube number, but not both.

6) Digit F is either an odd number or a triangle number, but not both.

7) Digit G is either an odd number or a prime number, but not both.

8) Digit H is either an even number or a square number, but not both.

9) Digit I is either a square number or a cube number, but not both.

10) Digit J is either a prime number or a triangle number, but not both.

11) A < B, C < D, E < F, G < H, I < J

12) A + B + C + D + E < F + G + H + I + J

Here is that MYSTERY number!


The FIRST Case of Mystery Number

The SECOND Case of Mystery Number 

Demystifying The Third Mystery Number

What was the challenge?

Taking a look at the given clues first for identifying the mystery number ABCDEFGHIJ.

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

1) Digit A is either a square number or a triangle number, but not both.

2) Digit B is either an even number or a cube number, but not both.

3) Digit C is either a cube number or a triangle number, but not both.

4) Digit D is either an odd number or a square number, but not both.

5) Digit E is either an odd number or a cube number, but not both.

6) Digit F is either an odd number or a triangle number, but not both.

7) Digit G is either an odd number or a prime number, but not both.

8) Digit H is either an even number or a square number, but not both.

9) Digit I is either a square number or a cube number, but not both.

10) Digit J is either a prime number or a triangle number, but not both.

11) A < B, C < D, E < F, G < H, I < J

12) A + B + C + D + E < F + G + H + I + J  

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

STEPS  :

Even numbers between 0 to 9 - 0, 2, 4, 6, 8

Odd numbers between 0 to 9 - 1, 3, 5, 7, 9

Square numbers between 0 to 9 - 0, 1, 4, 9 

Cube numbers between 0 to 9 - 8.

Prime numbers between 0 to 9 - 2, 3, 5, 7

Triangle numbers between 0 to 9 - 0, 1, 3, 6

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

STEP 1 : 

A can't be 0 as no leading 0's in the mystery number ABCDEFGHIJ.

Remember both of hints can't be true at the same time. That is Digit A is either a square or a triangle number but not both. Since, 0 and 1 are square as well as triangle numbers, A can't be 0 or 1. 

Similarly, 8 is even as well as cube, B can't take 8. And so on.

Possible values of A -  3, 4, 6, 9. (Square/Triangle)

Possible Values of B - 0, 2, 4, 6. (Even/Cube).

Possible Values of C - 0, 3, 6, 8. (Cube/Triangle)

Possible Values of D - 3, 4, 5, 7. (Odd/Square)

Possible Values of E - 3, 5, 7, 8. (Odd/Cube)

Possible Values of F - 0, 5, 6, 7, 9 (Odd/Triangle)

Possible Values of G - 1, 2, 9 (Odd/Prime)

Possible Values of H - 0, 2, 6, 8, 9 (Even/Square)

Possible Values of I -  4, 8, 9 (Square/Cube)

Possible Values of J -  0, 1, 2, 5, 6, 7  (Prime/Triangle) 

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

STEP 2 :

The value of I can't be 8 or 9 as there will be no digit left for J which is greater than I. 

Hence, I = 4.
 
--------------------------------------------------------------------------------- 

STEP 3 : 

Digit A can't be 6 or 9 since no digit will be left for B>A. 

As we have, I = 4, the digit A = 3 and hence B = 6.

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

STEP 4 :

The only digit left for C is 0, i.e. C = 0 as C can't be 8 for C<D to be true.

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

STEP 5 :

  Since I = 4 itself, so digit J can't be 0, 1, 2 for I<J to be true.

Digit D and digit J take either take 5 or 7. With that the only valid digit left for E is 8 to satisfy the condition E<F. 

Hence E = 8.

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

STEP 6 :

If E = 8, then F has to be 9. So, F = 9.
 
--------------------------------------------------------------------------------- 

STEP 7 : 

With 6, 8, 9 already taken up by other letters, the only digit left for H is 2 for G<H to be true. So, H = 3 and hence G = 1.

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

STEP 8 :

As per condition 12, 

A + B + C + D + E < F + G + H + I + J 

3 + 6 + 0 + D + 8 < 9 + 1 + 2 + 4 + J

17 + D < 16 + J 

As deduced in STEP 4, D must be 5 or 7 and J must be 7 or 5.

For above equation to be true, D = 5 and J = 7.


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

Conclusion :

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

Hence, the number ABCDEFGHIJ must be 3605891247.

Verifying the given hints - 

1) 3 is a triangle number, but not a square number.
2) 6 is an even number, but not a cube number.
3) 0 is a cube number, but not a triangle number.
4) 5 is an odd number, but not a square number.
5) 8 is a cube number, but not an odd number.
6) 9 is an odd number, but not a triangle number.
7) 1 is an odd number, but not a prime number.
8) 2 is an even number, but not a square number.
9) 4 is a square number, but not a cube number.
10) 7 is a prime number, but not a triangle number.
11) 3 < 6, 0 < 5, 8 < 9, 1 < 2, 4 < 7
12) 3 + 6 + 0 + 5 + 8 = 22, 9 + 1 + 2 + 4 + 7 = 23, 22 < 23  

Divide 1 Cube into 20 Cubes!

From a 1987 Hungarian math contest for 11-year-olds:

How can a 3 × 3 × 3 cube be divided into 20 cubes (not necessarily the same size)?

Cut this way to get 20 cubes....

Division of 1 Cube into 20 cubes

What was the challenge?

Mark cube for cutting 3 x 3 x 3 = 27 cubes. Cut any section of 2x2x2 = 8 cubes & cut rest of 27-8 = 19 cubes. So these 19 cubes plus 1 cube of 2x2x2 give us total number of 20 cubes. 

Cut The Blue Cube Puzzle

A solid, four-inch cube of wood is coated with blue paint on all six sides.

Then the cube is cut into smaller one-inch cubes. These new one-inch cubes will have either three blue sides, two blue sides, one blue side, or no blue sides. How many of each will there be?

Here is solution of the puzzle! 

Cut The Blue Cube Puzzle : Solution

What is the puzzle?

Apart from the 8 cubes at the center all 4 x 4 x 4 - 8 = 56 will have some paint on ones side at least. See below the 1/4 th cube is taken out.

The cubes at the 8 corners will have blue paint on three sides.

The cubes between corner cubes along 12 edges of big cube will have 2 sides painted. That is 12 x 2 = 24 cubes will painted with blue on 2 sides.

And 4 center cubes on each of 6 faces (left, right, top, bottom, front, back) will have only 1 side painted with blue. That is , there are 6 x 4 = 24 cubes having paint on one side only.

To conclude, out of 56 painted cubes,

24 cubes have paint on 1 side,

24 cubes painted with 2 sides,

8 are painted with three sides.

Unfair Game of Strange dice

Katherine and Zyan are playing a game using strange dice. Each die is a cube with six sides. Katherine's die has sides numbered 3, 3, 3, 3, 3, and 6. Zyan's die has sides numbered 2, 2, 2, 5, 5, and 5.

To play the game, Katherine and Zyan roll their dice at the same time and whoever rolls the higher value wins. If they play many times, who will win more frequently, Katherine or Zyan?

This person will be winning more! 
 

Advantage in Unfair Game of Strange Dice

What was the game?

Shortest Way : 

Katherine's die has sides numbered 3, 3, 3, 3, 3, and 6. And Zyan's die has sides numbered 2, 2, 2, 5, 5, and 5. That means whenever Zyan rolls a 2 then Katherine will always win. The probability that Zyan rolls 2 is 3/6 = 1/2. So in half of the cases, the Katherine will be winner of the game. Moreover, whenever, Zyan rolls a 5, Katherine can win if she rolls a 6. In short, in more than half cases, Katherine will win hence she has more advantage in this game.


Precise Way : 

Katherine will win if

1. Zyan rolls a 2 with probability 3/6 = 1/2

2. Zyan rolls a 5 (probability 3/6 = 1/2) and Katherine rolls a 6 (probability 1/6). So the probability for this win = 1/12.

Hence, Katherine has got 1/2 + 1/12 = 7/12 chances of winning.

Zyan will win if he rolls a 5 (probability 1/2) and Katherine rolls a 3 (probability 5/6) with probability = 1/2 x 5/6 = 5/12.  

That proves, Katherine has more chances (7/12 vs 5/12) of winning this game. 

Another Way : 

There can be 36 possible cases of numbers of cubes out of which in 15 cases Zyan seems to be winning and in 21 cases, Katherine is winning. Again, Katherine having more chances of winning (21/36 = 7/12) than Zyan (15/36 = 5/12).