Five identical safes are lined up in a row. Each safe requires a single numeric digit from 1 to 5 to open, and no two safes use the same digit. Safe A is immediately to the left of Safe B. Safe C is to the right of Safe B, but not immediately next to it. Safe D contains an even number, while Safe E contains the highest digit used among all the safes. Safe B's digit is exactly one less than Safe C's digit. What number opens Safe A?
2
Safe E is the highest digit (5). Safe C is to the right of Safe B, so the only remaining slots for them are 2, 3, or 4. Since Safe B's digit is one less than Safe C, they must be consecutive. The only consecutive pair left is 2 and 3 (with B=2, C=3) or 3 and 4 (with B=3, C=4). If B=3 and C=4, then A (which is to the left of B) could be 1 or 2. But B=2 and C=3 leaves A as 1 or 4. Let's test the constraints: Digits are 1, 2, 3, 4, 5. E=5. C=3, B=2. Since B is 2, A must be 1 or 4. Safe D is even (can be 4). If A=1, B=2, C=3, D=4, E=5, let's check Safe A being left of B (1 is left of 2), C is right of B (3 is right of 2, not immediately next to B? Wait, 3 is immediately next to 2. The clue says 'C is to the right of B, but not immediately next to it.' This means there must be a safe between B and C). Let's re-evaluate: If B=2 and C=4, there is a safe between them (Safe D=3). Then the sequence from left to right could be A, B, D, C, E. Let's check the digits: A=1, B=2, D=3, C=4, E=5. Let's check the clues: A is left of B (1 is left of 2). C is right of B, not immediately next to it (4 is separated from 2 by D). D contains an even number? Wait, 3 is odd. So D cannot be 3. Let's try another arrangement. B and C have a gap. Digits: 1, 2, 3, 4, 5. If B=1, C=3 (gap D=2). Let's check: B=1, C=3, gap D=2 (even). E=5. Remaining for A is 4. Let's check positions: A=4, B=1, D=2, C=3, E=5. Clue: A is left of B (4 is not left of 1). Let's rethink carefully: B and C are separated by one safe. The pairs with difference 1 are (1,2), (2,3), (3,4), (4,5). Since E=5, C cannot be 5 (otherwise B is 4, leaving no room for a safe between B and C). So C must be 4, meaning B is 3. Since there is a safe between B and C, that middle safe (let's say the 3rd safe) must hold the remaining even number, which is 2 (since D is even). So the digits in order are: A, B, [middle], C, E. We have digits 1, 3, 2, 4, 5 used for positions 1, 2, 3, 4, 5. The middle safe must be D=2. That leaves A=1. Let's check: A=1, B=3, D=2, C=4, E=5. Is A left of B? Yes (1 is left of 3). Is C right of B, not immediately next? Yes (4 is right of 3, separated by 2). Is D even? Yes (2 is even). Is E the highest? Yes (5). Therefore, Safe A opens with 1? Wait, let's re-read: 'Safe B's digit is exactly one less than Safe C's digit.' If B=3 and C=4, B is one less than C. Let's check the digits: A=1, B=3, D=2, C=4, E=5. Wait, the sequence of safes is A, B, [gap], C, E. That's 5 safes. The digits used are 1, 3, 2, 4, 5. Safe A's digit is 1? Wait, let's trace carefully: If B=2, C=3, the gap must hold a safe, say D=4. Then safes are A, B, D, C, E. Digits: A=1, B=2, D=4, C=3? No, C must be 3, which is adjacent to 2, violating 'not immediately next to it'. What if B=1, C=2? Then gap is 3. Safes: A, B, D, C, E -> A=4, B=1, D=3, C=2, E=5. Let's check: B=1 is one less than C=2. C is right of B, separated by D=3. D is odd, but D must be even ('Safe D contains an even number'). So D must be 2 or 4. If D=2, and C=3, then B=2? No. Let's solve systematically: B + 1 = C. E = 5. D is even (2 or 4). Safes: 1, 2, 3, 4, 5. Since C is right of B with a safe in between, the distance in indices is 2. So if B is at index 1, C is at index 3. The safes are A, B, [D/middle], C, E. Let's assign digits: E=5 (at index 5). C is at index 4, so C=4. Then B is at index 2, so B=3 (since B+1 = C? Wait, if C=4, B=3). Then the middle safe is index 3, which must be D. D must be even, so D=2. That leaves index 1 (Safe A) with the remaining digit, 1. Let's check: Safes are A(1), B(3), D(2), C(4), E(5). Positions: A is at 1, B is at 2. Is A immediately to the left of B? Yes, index 1 is immediately left of index 2. What is A's digit? 1. Wait, let's check the position of C: C is at index 4. B is at index 2. Is C to the right of B, not immediately next to it? Yes, index 4 is right of index 2, with index 3 between them. What is at index 3? Safe D, which has digit 2 (even). What is at index 5? Safe E, which has digit 5 (highest). What about B and C's digits? B=3, C=4. Is B's digit one less than C's? Yes, 3 is one less than 4. What digits are left for A? 1. Thus Safe A holds 1.
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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