Javob va yechimni ko‘rish
Javob: eng katta: 17, eng kichik: 6
Berilgan sonni $\overline{x6y5z}=100000x+6000+100y+50+z$ deb yozamiz.
11 ga bo‘linish shartidan $\overline{x6y5z}\equiv0\pmod{11}$ bo‘lsa, $100000\equiv1$, $6000\equiv6$, $100\equiv1$, $50\equiv6\pmod{11}$, shuning uchun
$\,x+6+y+6+z\equiv0\pmod{11}\;$ ya’ni $\,x+y+z\equiv -12\equiv10\pmod{11}$.
5 ga bo‘linish sharti esa $\overline{x6y5z}\equiv0\pmod{5}$, ya’ni oxirgi raqam $z$ $0$ yoki $5$ bo‘lishi kerak. $z=0$ yoki $5$ qo‘yib, $x+y\equiv10-z\pmod{11}$ va $x,y$ raqamlar $0\le x,y\le9$ bo‘lgani uchun mumkin bo‘lgan juftliklar:
$(x,y)=(9,6)$ → $x+y=15$, $(8,8)$ → $16$, $(7,9)$ → $16$, $(9,7)$ → $16$, $(8,9)$ → $17$, $(9,8)$ → $17$.
Shu bilan $x+y$ ning eng kichik qiymati $6$ (masalan $x=0,\;y=6,\;z=5$), eng katta qiymati $17$ (masalan $x=9,\;y=8,\;z=0$).
Demak, to‘g‘ri javob **B**: eng katta $17$, eng kichik $6$.