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.

The Numbered Hats Test!

One teacher decided to test three of his students, Frank, Gary and Henry. The teacher took three hats, wrote on each hat an integer number greater than 0, and put the hats on the heads of the students. Each student could see the numbers written on the hats of the other two students but not the number written on his own hat.

The teacher said that one of the numbers is sum of the other two and started asking the students:

— Frank, do you know the number on your hat?

— No, I don’t.

— Gary, do you know the number on your hat?

— No, I don’t.

— Henry, do you know the number on your hat?

— No, I don’t.

Then the teacher started another round of questioning:

— Frank, do you know the number on your hat?

— No, I don’t.

— Gary, do you know the number on your hat?

— No, I don’t.

— Henry, do you know the number on your hat?

— Yes, it is 144.

What were the numbers which the teacher wrote on the hats?

Here are the other numbers!

Source 

Cracking Down The Numbered Hats Test

What was the test?

Even before the teacher starts asking, the student must have realized 2 facts.

1. In order to identify numbers in this case, the numbers on the hats has to be in proportion i.e. multiples of other(s). Like if one has x then other must have 2x,3x etc.

2. Two hats can't have the same number say x as in that case third student can easily guess the own number as 2x since x-x = 0 is not allowed.

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Now, if numbers on the hats were distributed as x, 2x, 3x then the student wearing hat of number 3x would have quickly responded with correct guess. That's because he can see 2 number as x and 2x on others hats and he can conclude his number as x + 2x = 3x since 
2x - x = x is invalid combination (x, x, 2x) where 2 numbers are equal.

Other way, he can think that the student with hat 2x would have guessed own number correctly if I had x on my own hat. Hence, he may conclude that the number on his hat must be 3x.

But in the case, all responded negatively in the first round of questioning. So x, 2x, 3x combination is eliminated after first round.

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That means it could be x, 3x, 4x combination of numbers on the hats.

In second round of questioning, Henry guessed his number correctly.

If he had seen 3x and 4x on other 2 hats then he wouldn't have been sure with his number whether it is x or 7x.

Similarly, he must not have seen x and 4x as in that case as well he couldn't have concluded whether his number is either 5x or 3x.

But when he sees x and 3x on other hats he can tell that his number must be 4x as 2x (x,2x,3x combination) is eliminated in previous round!

So Henry can conclude that his number must be 4x.

Since, he said his number is 144,

4x = 144

x = 36

3x = 108.

Hence, the numbers are 36, 108, 144.

Develop an Unbeatable Strategy!

Consider a two player coin game where each player gets turn one by one. There is a row of even number of coins, and a player on his/her turn can pick a coin from any of the two corners of the row. The player that collects coins with more value wins the game. 

Develop a strategy for the player making the first turn, such he/she never looses the game.

Note : The strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners.

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Example :

  18 20 15 30 10 14

First Player picks 18, now row of coins is
  20 15 30 10 14

Second player picks 20, now row of coins is
  15 30 10 14

First Player picks 15, now row of coins is
  30 10 14

Second player picks 30, now row of coins is
  10 14

First Player picks 14, now row of coins is
  10 

Second player picks 10, game over.

The total value collected by second player is more (20 +
30 + 10) compared to first player (18 + 15 + 14).
So the second player wins.
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This is the unbeatable strategy! 

The Unbeatable Strategy for a Coin Game!

Why strategy needed in the case?

Let's recall the example given in the question.
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Example

  18 20 15 30 10 14

First Player picks 18, now row of coins is
  20 15 30 10 14

Second player picks 20, now row of coins is
  15 30 10 14

First Player picks 15, now row of coins is
  30 10 14

Second player picks 30, now row of coins is
  10 14

First Player picks 14, now row of coins is
  10 

Second player picks 10, game over.

The total value collected by second player is more (20 +
30 + 10) compared to first player (18 + 15 + 14).
So the second player wins.
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Here, it's very clear that the player who chooses coins numbered at even positions wins & the one who chooses odd position loses.

So first player who is going to choose coin first need to be smart. All that he/she need to do is make sum of values of all coin at even position, sum of values of all coins at odd positions and compare them. 

If he finds the sum of values of coins at odd position greater then he should choose 1st coin (odd position) followed by 3rd,5th.....(odd positions) & force other to choose even coins

And if he finds the sum of values of coins at even positions greater then he should choose
last coin (which is at even position as number of coins are even).

For example, in above case, 

  18 20 15 30 10 14

first player calculates 

Sum of odd coins = 18 + 15 + 10 = 43
Sum of even coins = 20 + 30 + 14 = 64.  

Since sum of even coins is greater, he should choose 6th coin (which will be followed by 4th and 2nd) and force other player to choose 5th,3rd and 1st coin.

Even in case second player selects 1st coin after 14 at other corner is selected by first, still he can be forced to choose the coin at 3rd position (odd position) if first player selects 2nd coin.

In short, first player need to make sure that he collects all the coins that are at even positions or odd position whichever has greater sum!   

Note that the total number of coins in the case can't be odd as then distribution of coins among 2 will be unequal.

Prove The Mathematical Fact

Show that the numbers from 1 to 15 can’t be divided into a group A of 13 numbers and a group B of 2 numbers so that the sum of the numbers in A equals the product of the numbers in B.

Here is the proof!

Proof of The Mathematical Fact!

What was that fact?

For a moment, let's assume that such group of 2 numbers exists whose product is equal to sum of rest 13 numbers taken out of 15 numbers.

Let x and y be those numbers in group B. Now x and y can be any number from 1 to 15. 

As per condition,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 - x - y = xy 

120 = xy + x + y 

Adding 1 to both sides,

121 = xy + x + y + 1

121 = x( y + 1 ) + 1( y + 1 )

121 = ( x + 1 ) ( y + 1 ) 

Since x & y are the numbers in between 1 to 15, possible value of x & y satisfying the above equation is 10. But x & y are must be 2 different number. Hence, our assumption goes wrong here!

So, the numbers from 1 to 15 can’t be divided into a group A of 13 numbers and a group B of 2 numbers so that the sum of the numbers in A equals the product of the numbers in B. 

What Could Be The Product?

Zach chooses five numbers from the set {1, 2, 3, 4, 5, 6, 7} and tells their product to Claudia. She finds that this is not enough information to tell whether the sum of Zach’s numbers is even or odd. What is the product that Zach tells Claudia?

Here is the way to get that product!

Guessing The Correct Product in Question!

What was the question?

When Zach tells the product of 5 numbers that he has chosen he indirectly conveying product of 2 un chosen numbers.

For example, product of all numbers in set = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040 and if Zach tell product 1 x 2 x 4 x 5 x 7 = 280 then obviously the product of numbers that he hasn't chosen is 5040/280 = 18 = 3 x 6.

Now, there are only 2 products viz. 12 and 6 which have more than 1 pair of numbers. 

The product 12 can be from pairs - (3,4) and (6,2)

The product 6 can be from pairs - (1,6) and (2,3) 

Here if the sum of un chosen numbers is odd (or even) then sum of other 5 chosen numbers also must be odd (or even).

In above cases, 6 has pairs whose sum is odd always and hence sum of other 5 numbers would be odd. In that case, Claudia would have been sure with if sum of numbers selected by Zach is either odd or even.

While in other case, the product 12 has 1 pair having sum odd (3,4) and other pair having sum even (6,2). Hence, Zach must have 'indirectly' suggested product 12 as a product of un chosen numbers that's why Claudia is saying that she doesn't know if the sum of numbers selected by Zach is even or odd.

Hence, the product of numbers selected by Zach = 5040 / 12 = 420.