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.

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.  

One More Alphamatic Problem?

In the following  puzzles, replace the same characters by the same numerals

so that the mathematical operations are correct.
               

Note - Each letter represents a unique digit and vice-versa.

ABCB - DEFC = GAFB

     :          +      -
  DH  x     AB =    IEI
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 GGE + DEBB = DHDG

Here is the SOLUTION 

One More Alphamatic Solution!

Look at the problem first!

Rewriting the problem once again,

ABCB - DEFC = GAFB
   :         +       -
  DH x   AB    =    IEI
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 GGE + DEBB = DHDG

We have 6 equations from above -

(1) A B C B - D E F C = G A F B  

(2) G G E + D E B B = D H D G  

(3) G A F B - I E I = D H D G 

(4) D E F C +  A B = D E B B 

(5) A B C B : D H  = G G E  

(6) D H x A B = I E I  

Steps :

1. From (1), we have B - C = B. That's possible only when C = 0.

2. If C = 0 then in (1), for tens' place subtraction i.e. C - F = F the carry need to 

    be taken from B. And that subtraction looks like 10 - F = F. Obviously, F = 5.

3. From (3), we see D in result seems to be carry and carry never exceeds 1 

    even if those numbers are 999 + 9999. So, D = 1. 

4. From (1), since C = 0, at hundreds' place (B - 1) - E = A and from (4),

    we have F + A = B (since first 2 digit of first numberremain same in result

    indicating no carry forwarded in addition of FC + AB = BB.

    So placing F = B - A in (B - 1) - E = A gives, F = E + 1. Since, F = 5, then E = 4. 

5. In (3), G at the thousands' place converted to D without actually subtraction 

    of digit from IEI. Since, G and D are different numbers some carry must have been

    taken from G.

    As D = 1 then G = 2.

6. From (1), A - D = G and D = 1 and G = 2 then A = 3 since if carry had been taken 

    from A then A = 4 which is impossible as we already have E = 4. 

7. From (2), E + B = G i.e. 4 + B = 2 only possible with B = 8.

8. With that, in (2), carry forwarded to  G + B = D making it 

    1 + G + B = 1 + 2 + 8 = 11 = 1D  i.e. carry 1 forwarded to G + E = H making it 

    1 + G + E = H = 1 + 2 + 4 = 7.

    Therefore, H = 7 and no carry forwarded as digit D in second number remains

    unchanged in result. 

9. Now (6) looks like - 17 x 38 = 646 = IEI = I4I. Hence, I = 6. 

To sum up,

A = 3, B = 8, C = 0, D = 1, E = 2, F = 5, G = 3, H = 7 and I = 6. 

Eventually, all above 6 equations after replacing digits in place of letters look - 

1. 3808 - 1450 = 2358  ✅

2. 224 + 1488 = 1712   ✅

3. 2358 - 646 = 1712   ✅

4. 1450 + 38 = 1488    ✅

5. 3808 : 17 = 224      ✅

6. 17 x 38 = 646         ✅  

Rewriting in the given format,

3808 - 1450 = 2358
      :         +       -
    17 x     38 =  646
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  224 + 1488 = 1712

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.