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"

Answer 1

$A=\bar{abcd}+\bar{dcba}=$
$1000a+100b+10c+d+1000d+100c+10b+a=$
$1001a+110b+110c+1001d=$
$1001(a+d)+110(b+c)=$
$11\cdot91(a+d)+11\cdot10(b+c)=$
$11[91(a+d)+10(b+c)]~\vdots~11$
$-----------------$
$121^{0}+1^{193}+(3^{2})^{3}:3^{5}=$
$1+1+3^{6}:3^{5}=1+1+3=5$
$-----------------$
$\{[(x+8):4+8\cdot9]:4+4\cdot5+9\}:25-1=1$
$\{[(x+8):4+72]:4+20+9\}:25=2$
$[(x+8):4+72]:4+29=50$
$[(x+8):4+72]:4=21$
$(x+8):4+72=84$
$(x+8):4=12$
$x+8=48$
$x=40$
$-----------------$
$a=2^{69}+2^{70}+2^{72}+2^{73}=$
$2^{69}(1+2+2^{3}+2^{4})=$
$2^{69}(1+2+8+16)=2^{69}\cdot27=2^{69}\cdot3^{3}$
$b=3^{49}+3^{48}-3^{47}-3^{46}=$
$3^{46}(3^{3}+3^{2}-3-1)=$
$3^{46}(27+9-3-1)=3^{46}\cdot32=3^{46}\cdot2^{5}$
$2^{69}\cdot3^{3}\Box 3^{46}\cdot2^{5}$
$2^{64}\Box 3^{43}$
$Mai~departe~nu~mai~stiu$