Agar seyfning kodi natural sonlardan iborat uchtalik bo'lsa, uni ochish uchun necha marta urinish kerak.

Agar seyfning kodi natural sonlardan iborat uchtalik bo'lsa, uni ochish uchun ne
  1. 10
  2. 8
  3. 9
  4. 5

📝 Detailed explanation

Seyf kodining oxirgi belgisi (Δ) noma'lum. Shartda aniq "natural sonlar" deyilgan.
Matematikada 0 raqami natural son hisoblanmaydi (natural sonlar 1 dan boshlanadi).
Demak, oxirgi xona uchun 1 dan 9 gacha bo'lgan jami 9 ta raqam mos keladi, ya'ni 9 marta urinish kifoya.
10 / 10 (100%)
Question 10 of 10
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