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## Mystery Number 8

Logic puzzles require you to think. You will have to be logical in your reasoning.

 Puzzle ID: #49152 Fun: (2.07) Difficulty: (3.08) Category: Logic Submitted By: cnmne

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) If A > B, then C = 5 or 7, else C = 0 or 1.
2) If B > C, then D = 1 or 2, else D = 4 or 9.
3) If C > D, then E = 6 or 9, else E = 3 or 5.
4) If D > E, then F = 2 or 4, else F = 1 or 6.
5) If E > F, then G = 5 or 6, else G = 0 or 7.
6) If F > G, then H = 1 or 4, else H = 8 or 9.
7) If G > H, then I = 0 or 8, else I = 6 or 7.
8) If H > I, then J = 3 or 8, else J = 2 or 5.
9) If I > J, then A = 3 or 7, else A = 4 or 8.
10) If J > A, then B = 0 or 9, else B = 2 or 3.

8274965103

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

1) A = 8 (I < J)
2) B = 2 (J < A)
3) C = 7 (A > B)
4) D = 4 (B < C)
5) E = 9 (C > D)
6) F = 6 (D < E)
7) G = 5 (E > F)
8) H = 1 (F > G)
9) I = 0 (G > H)
10) J = 3 (H > I)

Brute force methods can be used to arrive at a solution. However, process of elimination does work. For example, assign 8 to H. I can not exceed H, so J would have to be 3 (8 already assigned to H). G can not exceed H, so I would have to be either 6 or 7. I would be greater than J, so A would have to be 7 (3 already assigned to J), meaning I would have to be 6. J is less than A, so B would be 2 (3 already assigned to J). A is greater than B, so C would be 5 (7 already assigned to A). The only remaining numeral for G is 0, which means F would have to be greater than G, meaning H would have to be either 1 or 4. H has already been assigned 8, rendering that assignment impossible. Continue eliminating possibilities in the order shown below.

A) H can not be 8. C can not be 1. C can not be 0. B can not be 9. D can not be 1. D can not be 2. F can not be 1.
B) H must be 1. H can not be 4. H can not be 9.
C) G can not be 0. I can not be 6. I can not be 7. B can not be 0.
D) I must be 0. I can not be 8.
E) J can not be 2. J can not be 5. J can not be 8.
F) J must be 3. A can not be 3. B can not be 3. E can not be 3.
G) B must be 2. F can not be 2.
H) A must be 8. A can not be 4. A can not be 7.
I) D can not be 9.
J) E must be 9. E can not be 5. E can not be 6.
K) D must be 4. F can not be 4.
L) F must be 6. G can not be 6.
M) G must be 5. C can not be 5. G can not be 5.
N) C must be 7.

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 duchin38 Nov 28, 2011 I just did 0123456789, which is that number just not in that order....can that work? oddrey Apr 19, 2012 It's been a while since I took Computer Science, but is it just me or is this teaser written in Java code? dewtell Jan 27, 2015 Nice challenging teaser, with lots of possibilities to explore. Not sure if I follow your chain of possibilities to eliminate. I did it as follows: 1) Assume H=8, as you did. Leads to a contradiction, H can not be 8. 2) Assume H=9. This then implies FB, C>D, E=6, I=7, DF, G=5, and a contradiction, as that does not leave any legal values for C. So H can not be 9. 3) Therefore, F>G. G can not be 6 or 7. Assume that G=0 (the easiest way to get F>G). Then G F. E must be 9 to be > F. We must have C>D to have E be 9, so C must be 7, since it can not be 5. This implies A>B, and we know B dewtell Jan 27, 2015 Lost a bunch of the analysis in the posting there. Anyhow, the next bit was showing that either B or D must be 9 under the G=0 hypothesis, and both of them lead to contradictions. So F=6 and G=5. Now I must be 0 or 8, 8 leads to contradictions, so I=0, and we get to the given solutions. Eminem Jun 30, 2015 Hard enough but solvable saska Apr 30, 2017 The really simple scip model that I used to solve this teaser can be found at https://pastebin.com/GPk2xpLC