Question

sa se determine numerele naturale ab care indeplinesc conditia , unde a nu este egal b  :     a) aba + ab + ba = 732         

     b) a0a+b0b+aa+bb+a0+b0=854

c)aa0+bb0+aa+b+a0+b0=786

 

 

 

 

                    va rog ajutatimaaaaaaaaaa    :( :( :(

 

Answer 1

a) aba + ab + ba = 732

100*a+10*b+a+10*a+b+10*b+a=732
112*a+21*b= 732
obs: b trebuie sa fie par deci b=2,4,6 sau 8
daca b≤ 8 => a>5
dar a≤6
Avem 1 varianta
a=6=>dar 6*112+21b =672+21b= 732 => b=60:21...ceea ce nu reprezinta o solutie, deci exercitiul e gresit.


b) a0a+b0b+aa+bb+a0+b0=854
100a+a+100b+b+10a+a+10b+b+10a+10b=854
122a+122b=854
122*(a+b)=854
a+b=7
ab ∈(70,61,52,43,34,25,16)

c)aa0+bb0+aa+bb+a0+b0=786
100a+10a+100b+10b+10a+a+10b+b+10a+10b=786
131a+131b =786
131(a+b)=786
a+b=6
ab∈(60,51,42,24,15)