An old manuscript contains a four-digit code. The sum of all four digits is 10. The product of the first two digits equals the product of the last two digits, which is 6. The third digit is greater than the fourth. What is the code?
2331
Pairs that multiply to 6 are (1,6), (2,3), (3,2), (6,1). If the first two are 2 and 3, and the last two are 3 and 1, the sum is 2+3+3+1 = 9 (wait, 2+3+3+2 = 10). Let's check 2, 3, 3, 2: sum is 10, products are 6 and 6, but third digit isn't greater than fourth. If digits are 3, 2, 3, 2 sum is 10, products are 6 and 6. Let's look at digits 1, 6, 2, 3: sum is 12. Let's re-evaluate: digits 2, 3, 3, 2 has sum 10, products 6 and 6. Wait, the clue says 'third digit is greater than the fourth'. Let's use 3, 2, 4, 1? Products are 6 and 4 (no). What about 2, 3, 4, 1? Products 6 and 4. Ah, product of first two is 6 (e.g., 2 and 3) and product of last two is 6 (e.g., 3 and 2? No, 6 and 1). Code: 2, 3, 6, 1. Sum: 2+3+6+1 = 12. Let's use digits 1, 6, 3, 2. Sum: 1+6+3+2 = 12. What about 2, 3, 3, 2? Sum 10. Let's fix the riddle logic: First two digits multiply to 6 (2x3), last two digits multiply to 6 (3x2). Sum is 10. Wait, 2+3+3+2 = 10. Third is 3, fourth is 2. 3 > 2. Perfect!
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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