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. 

 

Count The Number of Arrangements

There are 10 parking spaces numbered from 101 to 110. At least one car is parked in these slots. If cars can be parked only at the consecutively numbered parking slots, how many such arrangements can be made. 

Consider that only one car can be parked in one parking slot and all cars are identical.

Here is the possible count! 

Possible Number of Arrangements

What was the puzzle?

Suppose there is only 1 car that is to be parked in 1 of the 10 slots. 

Number of possible arrangement = 10C1 = 10!/1!9 = 10.

That is 1 car can be parked in 10 slots in 10 number of ways.

Now, let's suppose that there are 2 cars that to be parked in 2 of the 10 parking slots. But the condition is that they need to be parked in consecutive slots. 

Among 10 slots for there are 9 possible consecutive slots for 2 cars. That is, 2 cars can be parked in consecutive slots in 9C1 = 9 number of ways. It's like placing 1 group of cars (having 2 cars) in 9 possible slots.

Similarly, in 10 parking slots for parking 3 cars there are 8 possible consecutive slots. Hence, there are 8 such arrangements are possible.

And so on for the rest number of cars.

Hence, there are total 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55 such arrangements are possible.  

Story of Farmer's 3 Sons

A farmer’s wife made some chapatis…The farmer had 3 sons…. 

The first son came, gave one chapati to the dog,and made three equal parts of remaining chapatis, ate one part of it and left the other two parts for his brothers…. Other two sons came one after the other and did the same thinking that they came first…

Then at night all three came to the house, one of them gave one chapati to dog and made three equal parts and the three brothers ate one-one part of it… 

If no chapati was broken in pieces then how many minimum number chapatis did the mom made?

Here are MATHEMATICAL steps for solution!