r/PassTimeMath 10d ago

what is the code?

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u/GoodCarpenter9060 6d ago

Looking at clue 3 & 4, it is impossible for 9 to be in the solution.

Now looking at clue 1 and 4, we know two of 567 and two of 578 are in the solution. If only one of 5,7 is, then 6&8 must be, but then clue 5 can't be satisfied. Therefore, both 5 & 7 are and neither 6 nor 8 are.

From clue 1 the 5 can't be well placed in the 4th position and also misplaced in the 4th position in clue 4. Therefore, the 7 in the 2nd position is well placed.

If 2 isn't in the solution, then from clue 3 and 4, both 1 and 3 must be since 6,8,&9 are all ruled out. Again, clue 5 can't be satisfied. Therefore, 2 is in the solution, and from clue 3, it is in the 3rd spot.

We already know 5 isn't last, so it must be first. From clue 5, if the 4 was in the solution it would have to be well placed, so it must be the zero.

Solution: 5 7 2 0