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

Avvalo faktorialning ta'rifini eslaymiz: $n! = 1\cdot2\cdot\ldots\cdot n$. Shunday qilib \[ \frac{25!}{23!}= \frac{1\cdot2\cdot\ldots\cdot23\cdot24\cdot25}{1\cdot2\cdot\ldots\cdot23}=24\cdot25 . \] Bu sonni tub ko‘paytuvchilarga ajratamiz: $24=2^{3}\cdot3$ va $25=5^{2}$, demak \[ 24\cdot25 = 2^{3}\cdot3\cdot5^{2}. \] Agar $n=p_{1}^{a_{1}}p_{2}^{a_{2}}\ldots p_{k}^{a_{k}}$ bo‘lsa, $n$ ning natural bo‘luvchilar soni \[ (a_{1}+1)(a_{2}+1)\ldots(a_{k}+1) \] ga teng bo‘ladi. Bizning holatda \[ (3+1)(1+1)(2+1)=4\cdot2\cdot3=24. \] Shunday qilib, $\frac{25!}{23!}$ sonining natural bo‘luvchilar soni **24** ga teng, ya’ni javob **B**.