Wait — perhaps we made a directional error. Let's suppose the number is three less than a multiple of each, but again, same.

February 22, 2026

Wait — perhaps we made a directional error. Let's suppose the number is **three less than a multiple of each**, but again, same.

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But for olympiad style, perhaps the intended setup was different. Let’s reframe: suppose the number is **three less than a multiple of 7 and 8**, but not necessarily 9? But question says of 7,8,9.
Alternatively, maybe the number is three less than a multiple of **the lcm**, but we already saw $ x = 501 $ is minimal.
But since the problem asks for a **two-digit** integer, and none exists, perhaps the intended answer is that **no such number exists**, but that’s not typical for olympiads.
Alternatively, suppose the question meant:
"A two-digit number that leaves remainder 3 when divided by 7, 8, and 9" — i.e., $ x \equiv 3 \pmod{7}, x \equiv 3 \pmod{8}, x \equiv 3 \pmod{9} $
Then $ x - 3 $ divisible by 7,8,9 → $ x - 3 = \text{lcm}(7,8,9) \cdot k = 504k $ → only $ x = 507 $ at $ k=1 $, still too big.

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