Question

Să se determine toate numerele naturale de forma ab , ştiind că :
a.) 1ab+1ba= 299 ; 
b.) a0b+b0a = 505 ; 
c.) aabb+bbaa=3333 ; 
d.) aba+bab=777.  
+_+ Cine stie rezolvarea acestui exercitiu ? 

Answer 1

a). 1ab + 1ba = 299
 100 + 10a +b +100 +10b +a = 299/-200
11a +11b = 99
11(a+b) = 99
a+b = 9  => a=0 b=9/ a=9 b=0/ a=1 b=8/b=8 a=1/ a=2 b=7 sia asa mai departe
b). a0b + b0a =505
100a + b + 100b +a =505
101a +101b=505
101(a+b) = 505
a+b = 5 => a=0 b=5/ a=5 b=0/ a=1 b=4/ a=4 b=1 ....
c).aabb+bbaa=3333
1000a +100a +10b +b +1000b+100b+10a+a = 3333
1111a+1111b=3333
1111(a+b)=3333
a+b= 3 => a=0 b=3/ a=3 b=0/ a=1 b=2/ b=1 a=2.....
d).aba+bab=777
100a+10b +a +100b +10a +b = 777
111a+111b =777
111(a+b)= 777
a+b=7 => a=0 b=7/a=7 b=0/ a=1 b=6/ a=6 b=1 .....\
Bafta!!

Answer 2

a) 1ab + 1ba = 299
100 + 10a +b +100 +10b +a = 299
11a+11b = 299-100
11a+11b = 99
11(a+b) = 99
a+b = 9  =>
 a=0  nu e bun
18, 27, 36,45,54,63,72,81,90

b) a0b + b0a =505
100a + b + 100b +a =505
101a +101b=505
101(a+b) = 505
a+b = 5
14,23,32,41,50
c)aabb+bbaa=3333
1000a +100a +10b +b +1000b+100b+10a+a =3333
1111a+1111b=3333
1111(a+b)=3333
a+b= 3
12,21,30
d)aba+bab=777
100a+10b +a +100b +10a +b = 777
111a+111b=777
111(a+b)=777
a+b=7 =>
16,25,34,43,52,61,70