Bo'linish belgilari, tub va murakkab sonlar bo‘yicha savol

48 sonini tub ko‘paytuvchilarga ajratamiz: $48=2^{4}\cdot 3^{1}$. Har bir tub ko‘paytuvchi $2$ ning $0,1,2,3,4$ darajalaridan va $3$ ning $0,1$ darajalaridan tanlanadi, shuning uchun bo‘luvchilar soni $(4+1)(1+1)=10$ ta bo‘ladi. Bu darajalarni birma‑bir qo‘shib, barcha bo‘luvchilarni olishimiz mumkin: \[ \begin{aligned} 2^{0}\cdot3^{0}&=1,\\ 2^{1}\cdot3^{0}&=2,\\ 2^{2}\cdot3^{0}&=4,\\ 2^{3}\cdot3^{0}&=8,\\ 2^{4}\cdot3^{0}&=16,\\ 2^{0}\cdot3^{1}&=3,\\ 2^{1}\cdot3^{1}&=6,\\ 2^{2}\cdot3^{1}&=12,\\ 2^{3}\cdot3^{1}&=24,\\ 2^{4}\cdot3^{1}&=48. \end{aligned} \] Natijada 48 ning barcha natural bo‘luvchilari: $1,2,3,4,6,8,12,16,24,48$. Bu ro‘yx A javobiga mos keladi.