Question

Arătat ca 11 | A , unde A=abcd barat+dcba barat
Va roog

Rezultatul calculului 121^0+1^193+(3^2)^3÷3^5 este....

{[(X+40:5):2^2+2^3ori3^2]:2^2+2^2ori5+3^2}:5^2-2^0=1

Dacă a=2^69+2^70+2^72+2^73 Si b=3^49+3^48-3^47-3^46 , atunci mai mare este numărul.....

,,^"-la puterea..... Ex:,, 3^49"
Cât mai repede va rog
Va dau coroana

Answer 1

A=abcd barat+dcba barat=1000a+100b+10c+d+1000d+100c+10b+a=
=1001a+110b+110c+1001d=11(91a+10b+10c+91d) divizibil cu 11

 121^0+1^193+(3^2)^3÷3^5 = 1+1+3^6÷3^5 =2+3=5

{[(x+40:5):2^2+2^3×3^2]:2^2+2^2×5+3^2}:5^2-2^0=1
{[(x+8):4+72]:4+20+9}:25-1=1
{[(x+8):4+72]:4+29}:25=2
[(x+8):4+72]:4=50-29
[(x+8):4+72]:4=21
(x+8):4+72=84
(x+8):4=12
x+8=48
x=40

a=2^69+2^70+2^72+2^73=2^69(1+2+2^2+2^3)=2^69×15=(2^3)^23×15=
=8^23×15
 
b=3^49+3^48-3^47-3^46 =3^46(3^3-3^2-3)=3^46×15=(3^2)^23×15=
=9^23×15

9>8
deci b>a