The Second 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) Either A = B / 3 or A = G + 3.

2) Either B = I - 4 or B = E + 4.

3) Either C = J + 2 or C = F * 3.

4) Either D = G * 4 or D = E / 3.

5) Either E = J - 1 or E = D / 4.

6) Either F = B * 2 or F = A - 4.

7) Either G = F + 1 or G = I - 3.

8) Either H = A / 2 or H = C * 3.

9) Either I = H + 3 or I = D / 2.

10) Either J = H - 2 or J = C * 2.

HERE is that MYSTERY Number! 

What was the FIRST case? 

Demystifying The Second Mystery Number

What was the challenge?

Let's take a look at clues given once again to identify number ABCDEFGHIJ.

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1) Either A = B / 3 or A = G + 3.

2) Either B = I - 4 or B = E + 4.

3) Either C = J + 2 or C = F * 3.

4) Either D = G * 4 or D = E / 3.

5) Either E = J - 1 or E = D / 4.

6) Either F = B * 2 or F = A - 4.

7) Either G = F + 1 or G = I - 3.

8) Either H = A / 2 or H = C * 3.

9) Either I = H + 3 or I = D / 2.

10) Either J = H - 2 or J = C * 2.

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STEPS :  

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STEP 1 : 

Few digits can be eliminated for few numbers straightaway on the first go. Those digits just don't 'fit' into the both equations provided for the particular letter.

1) Either A = B / 3 or A = G + 3.

    A - 0.

3) Either C = J + 2 or C = F * 3.

    C - 0, 1.

4) Either D = G * 4 or D = E / 3.

    D - 0, 5, 6, 7, 9

5) Either E = J - 1 or E = D / 4.

    E - 9 

6) Either F = B * 2 or F = A - 4.

    F - 7, 9

7) Either G = F + 1 or G = I - 3.

    G - 8 (since F can't be 7 & I can't be 11) 

8) Either H = A / 2 or H = C * 3.

    H - 0, 5, 7, 8 

9) Either I = H + 3 or I = D / 2.

    I - 8 (since H can't be 5 and obviously D can't be 16).

10) Either J = H - 2 or J = C * 2.

    J - 3 (since H can't be 5 & C can't be 1.5), 5 (since H can't be 7 & C can't 
    be  2.5), 9.

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STEP 2 :

Now, after eliminating some digits for letters we can revise the list of digits a particular letter(s) on LHS of equation can't make because the letter(s) on RHS doesn't (don't) take some digits.

For example, if J can't be 3 or 5 then C = J + 2 can't be 5 or 7. The deduction is supported by the other equation as F can't be 5/3 or 7/3. Hence, C can't be 5 or 7 for sure.

3) Either C = J + 2 or C = F * 3.

    C - 0, 1, 5, 7

4) Either D = G * 4 or D = E / 3.

    D - 0, 5, 6, 7, 9, 3 ( since E can't be 9 & G can't be 9/4).

5) Either E = J - 1 or E = D / 4.

    E - 9, 4 (J can't be 5 & D can't be 16), 8 (J can't be 9 & D can't be 32).

7) Either G = F + 1 or G = I - 3.

    G - 8, 0 [since G can't be -1 and I can't be 3 (since if I = 3 then I = H +3 
     gives H = 0 but H doesn't take 0)].

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STEP 3 : 

For B,

2) Either B = I - 4 or B = E + 4.

    B - 8 (since I can't be 12 & E can't be 4) 

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STEP 4 : 

Possible values left for D are -  1, 2, 4, 8. Remember, only one of the two given hints for the particular letter has to be true.
   
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      STEP 4.1 : 

       If D = 1, then D = E/3 (Hint 4) gives E = 3 & hence E = J - 1 (Hint 5) 
      gives J = 4. 

      Here, D = G*4 of Hint 4 must not be true as G would be 1/4.

      Similarly, E = D/4 of Hint 5 must be false and hence other hint i.e. 
      E = J - 1 must be true.

      Following the same logic as above -

      If D = 1 ---> E = 3 ----> E = J - 1 ---> J = 4 ----> 
      ---> J = H - 2 or J = C * 2.

      Hence, H = 6 or C = 2.

      If H = 6, then out of H = A / 2 or H = C * 3 only H = C*3 remains 
      valid which gives C = 2.

      And if C = 2, then C = J + 2 (Hint 3) gives J = 0.

      So, D = 1 produces 2 different values of J as 0 and 4. 

      Hence, this value of D is invalid. 

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     STEP 4.2 : 

     If D = 2, then (using Hint 4) E = 6 --> (Hint 5)---> J = 7  --->(Hint 10)

     ---> H = 9 ---> makes I = H + 3 invalid hence I = D/2 gives I = 1 

      I = 1 ---> leaves G = F + 1 (of Hint 7) valid.

     Further, H = 9 ---> (Hint 8) ---> C = 3 ---> (Hint 3) ---> 

      ---> F = 1 (since C = J + 2 = 9 invalid as H = 9 already) ---> (Hint 7)

      ---> G = 2.

     So if D = 2, the value of G will be also 2. That's against the rule.

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     STEP 4.3 :

     If D = 4, then (using Hint 4) G = 1 ----> (Hint 7) ---> 
     ---> F = 0 or I = 4 (invalid as D = 4 already) ---> (Hint 6) ---> A = 4. 

     Again, if D = 4, then A = 4 also hence this value of D is invalid.

     Hence, the only valid value of D is 8.

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STEP 5 : 

If D = 8, then using Hint 4 we have, G = 2  --->(Hint 7) ---> F = 1 or I =5

If I = 5, --->(Hint 9) ----> H = 2 but already we have G = 2. So,

G = 2  --->(Hint 7) ---> F = 1 ---> (Hint 6) ---> A = 5.

So far, we get,  D = 8, G = 2, F = 1, A = 5 at this stage so far.

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STEP 6 : 

Possible values left for E are - 0, 3, 6 and 7.
 
E = 0 ---> (Hint 5 ) ---> J = 1. But 1 is already taken by F.

E = 3 ---> (Hint 5 ) ---> J = 4 ---> (Hint 10) ---> H = 6 or invalid C = 2

H = 6 ---> (Hint 8 ) ---> invalid C = 2.

The digit 2 is already taken by G, so C = 2 is invalid. Hence, E = 3 is invalid.

E = 7 ---> (Hint 5 ) ---> J = 8. But 8 is already taken by G.

So, the only valid value of E = 6.

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STEP 7 :

E = 6 ---> (Hint 5 ) ---> J = 7 ---> (Hint 10) ---> H = 9 ---> (Hint 5) --->

H = 9 ---> (Hint 8) ---> C = 3

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STEP 8 :

The values we got so far,  E = 6, J = 7, H = 9, C = 3, D = 8, G = 2, F = 1, 
A = 5.

Only letters left are B and I while digits left are 0 and 4.

Correct hint out of "Either B = I - 4 or B = E + 4" must be B = I - 4 since 
B = E + 4 = 6 + 4 = 10 must be invalid.

So, B = I - 4 gives us I = 4 and B = 0 .

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Conclusion :

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

all together gives us number

ABCDEFGHIJ as 5038612947.

Let's verify the above number as per given hints.

1) A = G + 3 = 2 + 3 = 5.
2) B = I - 4 = 4 - 4 = 0.
3) C = F * 3 = 1 * 3 = 3.
4) D = G * 4 = 2 * 4 = 8.
5) E = J - 1 = 7 - 1 = 6.
6) F = A - 4 = 5 - 4 = 1.
7) G = F + 1 = 1 + 1 = 2.
8) H = C * 3 = 3 * 3 = 9.
9) I = D / 2 = 8 / 2 = 4.
10) J = H - 2 = 9 - 2 = 7.