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 Fearsome Logical Challenge

You and your two friends Pip and Blossom are captured by an evil gang of logicians. In order to gain your freedom, the gang’s chief, Kurt, sets you this fearsome challenge.

The three of you are put in adjacent cells. In each cell is a quantity of apples. Each of you can count the number of apples in your own cell, but not in anyone else’s. You are told that each cell has at least one apple, and at most nine apples, and no two cells have the same number of apples.

The rules of the challenge are as follows: 

The three of you will ask Kurt a single question each, which he will answer truthfully ‘Yes’ or ‘No’. Every one hears the questions and the answers. He will free you only if one of you tells him the total number of apples in all the cells.

    Pip: Is the total an even number?

    Kurt: No.

    Blossom: Is the total a prime number?

    Kurt: No

You have five apples in your cell. What question will you ask?

THIS should be the question that you need to ask!

Logical Response to The Fearsome Challenge

What was the challenge?

Remember, all you have to do is that ask one crucial question to logicians and not necessarily deduce the total count of apples.

Since, each cell has 1 to 9 apples and no two cells have same number of apples, the lowest count of apple is 1 + 2 + 3 = 6 and the highest count would be 7 + 8 + 9 = 24.

That is the total number of apples could be between 6 to 24.

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Now, Pip and Blossom already have gathered some information about the total.

1. The total is not an even number - Hence, only numbers  7,9,11,13,14,15,17,19,21,23 can represent the total count.

2. The Total is not a prime number - Out of the number above, only 9, 15, 24 are non-prime number.

Hence, the total count must be among 9,15 or 24.

Now, your task is easier. All you need to ask the Kert below question -

"Is total is 15?"
 
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CASE 1 : Total is really 15 -

Then Kert would reply with YES to your question and all of you know the total now.

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CASE 2 : Total is 9 -

The Kert's answer to your question would be NO.

If the total is 9 and you have 5 apples then rest of 4 apples must be distributed among Pip and blossom as (1,3) or (3,1) but can't be (2,2) since no 2 cells can have same number of apples.

Now, the friend having 1 apple (or 3 apples) can think that the total can't be 21 as in that case other 2 must have total of 20 (or 18) apples. But the maximum that other two can have is 9 + 8 = 17 apples.

So any of them can deduce that the total is 9.

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CASE 3 : Total is 21 -

Since you have 5 apple other 2 must be having total of 16 apples. One of them must be having 7 apples and other having 9 apples.

The friend having 9 apples can easily deduce the count at 21 since 9 as a total count is impossible in the case as for that the other must have 0 apples.

And the friend with 7 apples know that other can't have 1 + 1 or 2 + 0 (as per given data) apples in order to have total count of 9. Hence, he too can deduce that the total must be 21.

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To conclude, depending on the what Kert answers to your question and the count of apples that each of other 2 friends have one of them (or you too if count is 15) can deduce the total number of apples easily. And eventually, logicians have to set you free as promised.

Constructing Magical Square Using Prime Numbers

Whether it’s possible to construct a magic square using the first nine prime numbers (here counting 1 as prime):

1 2 3 5 7 11 13 17 19

Is it?

Find the possibility here!

Impossible Magical Square

What was the task given?

That's impossible task. All the listed prime numbers sums together to 78. For square to be magic, sum of each row & column must be equal. In this case, it should be 78/3 = 26.

For sum of 3 to be even, 1 must be even & other 2 odd (or all even). All 3 odd can't sum even.

In listed prime numbers there is only 1 even number i.e.2. Hence, for other 2 rows/columns we can't have even sum.