Mystery Number 7
Logic puzzles require you to think. You will have to be logical in your reasoning.
There is a tendigit 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.
Answer
5038612947
A = 5, B = 0, C = 3, D = 8, E = 6, F = 1, G = 2, H = 9, I = 4, J = 7
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.
A) Check each letter for any value that it can not be on the left side of an equation. For example, A can not be 0, as B would also have to be 0 or G would have to be negative. On the first pass through the letters, the following can be determined: A not 0, C not 0, C not 1, D not 0, D not 5, D not 6, D not 7, D not 9, E not 9, F not 7, F not 9, G not 8 (because F not 7), H not 0, H not 5, H not 7, H not 8, I not 0, I not 3 (because H not 0 and D not 6), I not 8 (because H not 5), J not 3 (because H not 5), J not 5 (because H not 7), and J not 9.
B) Make a second pass through the letters. The following can be determined: C not 5, C not 7, D not 3, E not 4, E not 8, and G not 0.
C) Make a third pass. The following can be determined: B not 8.
D) For D, only four numbers are still valid: 1, 2, 4, and 8. Plug each of these numbers into D and check the resulting flow of equations. If D = 1, then the resulting flow will lead to J having to be both 4 and 0. If D = 2, then the resulting flow will lead to 2 also being assigned to G. If D = 4, then the resulting flow will lead to 4 also being assigned to either A or I. If D = 8, then G = 2, F = 1, and A = 5.
E) For E, four numbers are still valid: 0, 3, 6, and 7. Plug each number into E and check the resulting flow. If E = 0, then the flow will lead to J being assigned 1, which has already been assigned to F. If E = 3, then the flow will lead to C being assigned 2, which has already been assigned to G. If E = 7, then the flow will lead to J being assigned 8, which has already been assigned to D. If E = 6, then J = 7, H = 9, and C = 3.
F) The only remaining value for I is 4, which leaves 0 being assigned to B.
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Comments
jhosek
Nov 24, 2010
 I got it. A little tricky but solvable. 
quesadillaman12
Jan 09, 2011
 My brain hurts. 
affanak
May 10, 2011
 Equation 1. A= B/3
From your solution:
B=0, which would imply that A=0 too.
0/3=3 
affanak
May 10, 2011
 Take back my comment (cant delete it) 
galanix
Feb 03, 2012
 Good puzzle. I followed a strategy similar to the one you displayed in the answer but slightly different. After a few passes instead of testing the different values of D I just tested which letter could be 0. At that point I had B,E,F, and J which could possibly be 0. I tried B as zero first and it worked. 
galanix
Feb 03, 2012
 Simplified version of the answer:
A = G + 3 = 5
B = I  4 = 0
C = F * 3 = 3
D = G * 4 = 8
E = J  1 = 6
F = A  4 = 1
G = F + 1 = 2
H = C * 3 = 9
I = D / 2 = 4
J = H  2 = 7 
sarggames
Oct 24, 2014
 EASY ONE I DID IT WITHOUT ANY FORCED THINKING 
javaguru
Nov 10, 2015
 I took a different approach by looking for rules that could be eliminated.
1) D = G*4 can be rewritten as G = D/4, so D = G*4 and E = D/4 can't both be true. Likewise D = E/3 can be rewritten as E = D*3, so D = E/3 and E = D/4 can't both be true. Since either D = G*4 or D = E/3 must be true, E = D/4 can't be true and E = J1 must be true.
2) J = C*2 can be rewritten as C = J/2, so J = C*2 and C = J+2 can't both be true. J = H2 can be rewritten as H = J+2, so J = C*2 and C = J+2 can't both be true. Since either J = C*2 or J = H2 must be true, C = J+2 can't be true and C = F*3 must be true.
(continued next comment because this exceeded make comment size) 
javaguru
Nov 10, 2015
 3) Looking at the rules for F, A & G:
F = B*2 or f = A4
G = F+1 or g = I3
A = B/3 or a = G+3
there are 8 combinations possible for a set of consistent rules (FGA, FgA, FGa, Fga, fGA, fgA, fGa, fga), some combinations can be eliminated immediately. g = I3 can be rewritten as I = g+3, so g = I3 and a = G+3 can't both be true, so combinations with ag are eliminated. f = A4 requires A to be at least 4 and A = B/3 requires A to be at most 3, so these rules can't both be true and combinations with fA are eliminated. This leaves (FGA, FgA, FGa, fGa).
FGA: F = B*2, G = F+1, A = B/3
In this case B must be at least 3, making F at least 6, but C = F*3 requires F to be at most 3. So FGA is eliminated.
FgA: F = B*2, g = I3, A = B/3
In this case B must be 3, so B = E+4 is eliminated and B = I4 is required, making I = 7. If I = 7, then I = D/2 is eliminated and I = H+3 is required. I = g+3 and I = H+3 can't both be true, so FgA is eliminated.
FGa: F = B*2, G = F+1, a = G+3
In this case a = F+4, so B must be 1 or 2, so B = E+4 is eliminated and B = I4 is required, making I = 5 or 6. I = D/2 requires I to be at most 4, so I = H+3 is required, making H = 2 or 3. If H = 2, then H = C*3 is eliminated and H = A/2 is required, making A = 4, but A = 4, makes F = 0 and B = 0, so H can't be 2. If H = 3, then H = A/2 is eliminated and H = C*3 is required, which makes C = 1, but C = F*3, so C can't be 1. FGa is eliminated.
fGa: f = A4, G = F+1, a = G+3
f = A4 and a = F+4, so these equations are consistent. fGa is the only choice.
(continued next comment) 
javaguru
Nov 10, 2015
 4) Now the following rules are established:
A = G+3
C = F*3
E = J1
F = A4
G = F+1
The equations for H are H = C*3 or H = A/2. For H = C*3 to be true, F = 1, C = 3 and H = 9. This makes A = 5 and G = 2. So far, so good. I = H+3 is eliminated, requiring I = D/2. D is either D = G*4 = 2*4 = 8 or D = E/3, requiring E = 6 and D = 2. If D = 8, then I = 4; if D = 2 then I = 1. I can't be 1, so if H = C*3, then D = G*4 = 8 and I = D/2 = 4, still good. B, E and J are left and 0, 6 & 7. Since E = J1, E = 6 and J = 7, leaving B = 0. J = H2 = 92 = 7 and B = I4 = 44 = 0.
So we have a solution that fits, although we haven't determined if it is the only solution.
A = G+3 = 2+3 = 5
B = I4 = 44 = 0
C = F*3 = 1*3 = 3
D = G*4 = 2*4 = 8
E = J1 = 71 = 6
F = A4 = 51 = 1
G = F+1 = 1+1 = 2
H = C*3 = 3*3 = 9
I = D/2 = 8/2 = 4
J = H2 = 92 = 7 
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