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

$125!$ ning $5$ ga bo‘linish darajasini topamiz. Har bir $5$ ning darajasi $5^k$ ga teng bo‘lishi uchun $125!$ ichida $5$ ning $k$ marta ko‘paytirilgan bo‘lishi kerak. Bu darajani quyidagicha hisoblaymiz: \[ \begin{aligned} \left\lfloor\frac{125}{5}\right\rfloor &= 25 \qquad\text{(5 ga bo‘linadigan sonlar)}\\ \left\lfloor\frac{125}{5^{2}}\right\rfloor &= \left\lfloor\frac{125}{25}\right\rfloor = 5 \qquad\text{(25 ga bo‘linadigan sonlar)}\\ \left\lfloor\frac{125}{5^{3}}\right\rfloor &= \left\lfloor\frac{125}{125}\right\rfloor = 1 \qquad\text{(125 ga bo‘linadigan son)}\\ \left\lfloor\frac{125}{5^{4}}\right\rfloor &= 0 \qquad\text{(keyingi darajalar yo‘q).} \end{aligned} \] Endi bu qiymatlarni qo‘shamiz: \[ 25+5+1 = 31. \] Shunday qilib, $1\cdot2\cdot3\cdots124\cdot125 =125!$ eng katta $5^{31}$ ga bo‘linadi. Javob: **C) 31**.