Bo'linuvchanlik. Sonning natural bo'luvchilar soni va yig'indisi bo‘yicha savol

Berilgan ifodani soddalashtiramiz: \[ \frac{8!}{6!}= \frac{8\cdot7\cdot6!}{6!}=8\cdot7=56. \] Endi $56$ sonining tub ko‘paytmalarini aniqlaymiz: \[ 56=2^{3}\cdot7^{1}. \] Biror sonning tub ko‘paytmalar ko‘rinishi $p_{1}^{a_{1}}p_{2}^{a_{2}}\dots p_{k}^{a_{k}}$ bo‘lsa, uning natural bo‘luvchilari soni \[ (a_{1}+1)(a_{2}+1)\dots (a_{k}+1) \] bo‘ladi. Shuning uchun \[ \text{bo‘luvchilar soni}= (3+1)(1+1)=4\cdot2=8. \] Demak, $\frac{8!}{6!}$ sonining natural bo‘luvchilari soni **8** ga teng. Javob: **C**.