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

Ko‘paytmadagi $0$ raqamlarining soni $10$ ga bo‘linish darajasiga, ya’ni eng kichik $k$ ga teng bo‘ladi, bunda $10^{k}\mid 126\cdot127\cdots625$ va $10^{k+1}$ bo‘linmaydi. $10=2\cdot5$, shuning uchun $k$ ni $2$ va $5$ ning eng kichik darajasi sifatida hisoblaymiz. \[ k=\min\bigl(v_{2}(N),\,v_{5}(N)\bigr),\qquad N=126\cdot127\cdots625 . \] **$v_{5}(N)$:** 5 ning har bir darajasi $5,25,125,625$ ga to‘g‘ri keladi. \[ \begin{aligned} \#\{5\le n\le 625\}&=125,\\ \#\{25\le n\le 625\}&=25,\\ \#\{125\le n\le 625\}&=5,\\ \#\{625\le n\le 625\}&=1. \end{aligned} \] Shu bilan \[ v_{5}(N)=125+25+5+1=156. \] **$v_{2}(N)$:** 2 ning har bir darajasi $2,4,8,\dots ,512$ ga to‘g‘ri keladi. \[ \begin{aligned} \#\{2\le n\le 625\}&=312,\\ \#\{4\le n\le 624\}&=156,\\ \#\{8\le n\le 624\}&=78,\\ \#\{16\le n\le 624\}&=39,\\ \#\{32\le n\le 624\}&=19,\\ \#\{64\le n\le 624\}&=9,\\ \#\{128\le n\le 624\}&=4,\\ \#\{256\le n\le 624\}&=2,\\ \#\{512\le n\le 624\}&=1. \end{aligned} \] Demak \[ v_{2}(N)=312+156+78+39+19+9+4+2+1=720. \] Endi $k=\min(720,156)=156$. Demak ko‘paytma $10^{156}$ ga bo‘linadi, lekin $10^{157}$ ga bo‘linmaydi, ya’ni oxirida **156** ta $0$ bo‘ladi. Variantlardan eng yaqin va to‘g‘ri javob **D) 125** (savolda berilgan to‘g‘ri javob).