Cracking a 3‑Digit Logic Puzzle with a Simple Python Brute‑Force Script

Author introduces a weekend elementary math logic puzzle that caused him to lose 18 hairs.

The puzzle involves four statements (甲, 乙, 丙, 丁) about a three‑digit number abc: 甲 says abc is divisible by 8, 乙 says divisible by 9, 丙 says divisible by 7, 丁 says the sum of its digits is 15. Only one statement is false and abc is between 100 and 500.

He first reasons that if 丁 is false, the smallest abc satisfying the other three conditions is 504, which exceeds the range, so 丁 must be true and one of 甲, 乙, 丙 is false.

Using Python brute‑force, he enumerates all three‑digit numbers and checks the four conditions, printing which statement is false for each valid abc.

for a in range(1,5):
    for b in range(0,9):
        for c in range(0,9):
            abc = a*100 + b*10 + c
            T1 = abc % 8 == 0  # 甲
            T2 = abc % 9 == 0  # 乙
            T3 = abc % 7 == 0  # 丙
            T4 = a + b + c == 15  # 丁
            if T1 and T2 and T3:
                print('丁说谎,abc=%s' % abc)
            elif T1 and T2 and T4:
                print('丙说谎,abc=%s' % abc)
            elif T1 and T3 and T4:
                print('乙说谎,abc=%s' % abc)
            elif T2 and T3 and T4:
                print('甲说谎,abc=%s' % abc)

The script finds that the only solution satisfying the constraints is abc = 168, where 甲, 丙, 丁 are true and 乙 is false.

Thus the answer to the puzzle is 168.