Question

pllsss ii urgent va roggg

Answer 1

fie abc primul numar, atunci al doilea numar va fi cba,

iar al treilea va fi 2(abc+cba).

deci, abc+cba+2(abc+cba)=1815 ⇔ abc+cba+2abc+2cba=

3(abc+cba)=1815 |:3 ⇔ abc+cba=605 ⇔

(100a+10b+c)+(100c+10b+a)=101a+101c+20b=605⇔

101(a+c)+20b=605, cum 5|20b, 5|605 ⇒ 5|[101(a+c)], gcd(5,101)=1 ⇒

5|(a+c) ⇒ (a+c)∈{5,10,15}, pentru ca 2≤a+c≤18, a,b,c cifre,

dar 101(a+c)≤605 ⇒ (a+c)≤6 ⇒ a+c=5.

deci, 101·5+20b=605 ⇒ 20b=605-505=100 ⇒ b=5.

5=1+4=4+1=2+3=3+2=a+c.

deci, abc∈{154, 451, 253, 352}.