A password consists of four different non-zero digits. The sum of the first two digits equals the sum of the last two digits. The first digit is three times the fourth digit. The second digit is an even number greater than 5. What is the password?
9623
Let digits be A, B, C, D. Second digit B is an even number > 5, so B = 6 or 8. First digit A = 3 × D. Since digits are 1-9, D can be 1, 2, or 3. If D = 3, A = 9. So A = 9, D = 3. We have 9, B, C, 3. Sum of first two = 9 + B. Sum of last two = C + 3. So 9 + B = C + 3, which means C - B = 6. Since B can be 6 or 8: if B = 8, C = 14 (impossible, digits 1-9). If B = 6, C = 12 (impossible). Wait! Let's re-read: second digit is even > 5, so B = 6 or 8. Wait, if B = 6, C - 6 = 6 => C = 12 (invalid). Let's re-evaluate D and A. What if D = 2? A = 6. Then A = 6, B can be 8. If A = 6, B = 8, sum = 14. Last two sum to 14. If D = 2, C + 2 = 14 => C = 12 (invalid). Let's check D = 1, A = 3. First two sum = 3 + B. B is even > 5 (6 or 8). If B = 6, sum = 9. Last two sum = C + D = 9. Since D = 1, C = 8. Digits: A=3, B=6, C=8, D=1. Let's check sum: 3 + 6 = 9, 8 + 1 = 9. First digit (3) is 3 times fourth digit (1). Second digit (6) is even and > 5. All digits are different (3, 6, 8, 1). The password is 3681.
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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