An ancient vault requires a four-digit numerical code. - The sum of all four digits is 10. - The first digit is twice the third digit. - The second digit is zero. - The last digit is odd and greater than 5. What is the four-digit code?
4024
Let the code be ABCD. We know B = 0. D is odd and > 5, so D must be 7 or 9. If D = 9, sum is A + 0 + C + 9 = 10, so A + C = 1. Since A = 2C, 3C = 1 (not an integer). Therefore, D must be 7? Wait, sum is A + 0 + C + D = 10. If D = 7, A + C = 3. Since A = 2C, 3C = 3, so C = 1 and A = 2. Let's check: A=2, B=0, C=1, D=7 -> 2017. Sum = 2+0+1+7 = 10. A (2) is twice C (1). B = 0. D (7) is odd and > 5. Code is 2017.
About this riddle
Logic riddles give you a few facts and ask what must be true. List what you know, cross out what cannot be, and see what is left.
Difficulty: Hard. More like this: Logic riddles with answers or try Einstein's Logic Grid for a different kind of puzzle.
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