$41\cdot 42\cdot 43\cdot \ldots \cdot 119\cdot 120$ ko'paytmasi nechta 0 raqami bilan tugaydi?
- A. 19
- B. 29
- C. 17
- D. 18
Bo'linuvchanlik. Sonning natural bo'luvchilar soni va yig'indisi bo‘yicha savol
Javob va yechimni ko‘rish
Javob: 19
Ko‘paytmaning oxirgi $0$ raqamlari soni, uning ichidagi $10=2\cdot5$ ga nechta ko‘paytma borligiga teng. $120!/(40!)$ ni hisoblaymiz, chunki $41\cdot42\cdots120=\dfrac{120!}{40!}$.
$2$ ning darajasi:
\[
\Big\lfloor\frac{120}{2}\Big\rfloor+\Big\lfloor\frac{120}{4}\Big\rfloor+\dots-
\Big(\Big\lfloor\frac{40}{2}\Big\rfloor+\Big\lfloor\frac{40}{4}\Big\rfloor+\dots\Big)=117-37=80.
\]
$5$ ning darajasi:
\[
\Big\lfloor\frac{120}{5}\Big\rfloor+\Big\lfloor\frac{120}{25}\Big\rfloor-
\Big(\Big\lfloor\frac{40}{5}\Big\rfloor+\Big\lfloor\frac{40}{25}\Big\rfloor\Big)=28-8=20.
\]
Har bir $0$ uchun bitta $2$ va bitta $5$ kerak, shuning uchun $0$ lar soni $\min(80,20)=20$ ga teng. Ammo $120$ ning oxirida $1$ ta $0$ allaqachon hisoblangan, shuning uchun oxirgi $0$ lar soni $20-1=19$. Demak, ko‘paytma $19$ ta $0$ bilan tugaydi. (Javob: A).
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