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Mystery Number 7

Category:Logic
Submitted By:cnmne
Fun:*** (2.27)
Difficulty:**** (3.16)



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.

Show Answer



Comments on this teaser


Posted by jhosek11/24/10
I got it. A little tricky but solvable.

Posted by quesadillaman1201/09/11
My brain hurts. :-?

Posted by affanak05/10/11
Equation 1. A= B/3 From your solution: B=0, which would imply that A=0 too. 0/3=3

Posted by affanak05/10/11
Take back my comment (cant delete it)

Posted by galanix02/03/12
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.

Posted by galanix02/03/12
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

Posted by sarggames10/24/14
:D EASY ONE I DID IT WITHOUT ANY FORCED THINKING

Posted by javaguru11/10/15
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 = J-1 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 = H-2 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 = H-2 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)

Posted by javaguru11/10/15
3) Looking at the rules for F, A & G: F = B*2 or f = A-4 G = F+1 or g = I-3 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 = I-3 can be rewritten as I = g+3, so g = I-3 and a = G+3 can't both be true, so combinations with ag are eliminated. f = A-4 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 = I-3, A = B/3 In this case B must be 3, so B = E+4 is eliminated and B = I-4 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 = I-4 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 = A-4, G = F+1, a = G+3 f = A-4 and a = F+4, so these equations are consistent. fGa is the only choice. (continued next comment)

Posted by javaguru11/10/15
4) Now the following rules are established: A = G+3 C = F*3 E = J-1 F = A-4 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 = J-1, E = 6 and J = 7, leaving B = 0. J = H-2 = 9-2 = 7 and B = I-4 = 4-4 = 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 = I-4 = 4-4 = 0 C = F*3 = 1*3 = 3 D = G*4 = 2*4 = 8 E = J-1 = 7-1 = 6 F = A-4 = 5-1 = 1 G = F+1 = 1+1 = 2 H = C*3 = 3*3 = 9 I = D/2 = 8/2 = 4 J = H-2 = 9-2 = 7




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