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

Ko‘paytma $1\cdot2\cdot3\cdots63\cdot64$ – bu $64!$ ga teng. $2$ ning eng katta darajasi $k$ ga bo‘linishini topish uchun $64!$ ichidagi $2$ ning umumiy ko‘paytuvchilar sonini hisoblaymiz (Legendre formulasi): \[ k=\Big\lfloor\frac{64}{2}\Big\rfloor+\Big\lfloor\frac{64}{2^{2}}\Big\rfloor+\Big\lfloor\frac{64}{2^{3}}\Big\rfloor+\Big\lfloor\frac{64}{2^{4}}\Big\rfloor+\Big\lfloor\frac{64}{2^{5}}\Big\rfloor+\Big\lfloor\frac{64}{2^{6}}\Big\rfloor. \] Hisoblaymiz: \[ \frac{64}{2}=32,\qquad \frac{64}{4}=16,\qquad \frac{64}{8}=8,\qquad \frac{64}{16}=4,\qquad \frac{64}{32}=2,\qquad \frac{64}{64}=1. \] Barchasini yig‘indisi: \[ k=32+16+8+4+2+1=63. \] Demak, $1\cdot2\cdot3\cdots64$ ko‘paytmasi $2^{63}$ ga bo‘linadi, lekin $2^{64}$ ga bo‘linmaydi. Eng katta daraja $63$, ya’ni javob **B**.