$1\cdot 2\cdot 3\cdot \ldots \cdot 120\cdot 121$ ko'paytma 11 ning qanday eng katta darajasiga bo'linadi?
- A. 14
- B. 13
- C. 16
- D. 12
Bo'linuvchanlik. Sonning natural bo'luvchilar soni va yig'indisi bo‘yicha savol
Javob va yechimni ko‘rish
Javob: 12
Ko‘paytma $1\cdot2\cdot3\cdots121$ ga $11$ ning nechta faktor sifatida kelishini aniqlash uchun $121!$ dagi $11$ ning darajasi $v_{11}(121!)$ ni hisoblaymiz. Legendre formulasiga ko‘ra
\[
v_{11}(121!)=\Big\lfloor\frac{121}{11}\Big\rfloor+\Big\lfloor\frac{121}{11^{2}}\Big\rfloor
= \lfloor 11\rfloor+\lfloor 1\rfloor=11+1=12 .
\]
Demak $1\cdot2\cdot\ldots\cdot121$ eng ko‘p $11^{12}$ ga bo‘linadi, ya’ni $11$ ning $12$-chi darajasiga. Javob: **D) 12**.
Mavzuni mustahkamlang
Bu savol Bo'linuvchanlik. Sonning natural bo'luvchilar soni va yig'indisi mavzusiga tegishli. Ta’riflar, formulalar va misollarni mavzu sahifasida ko‘ring.
Mavzuni o‘rganish