Question

arata ca 101|(a0b+b0a) oricare ar fi cifrele nenula a si b​

Answer 1

a0b(in baza 10) = 100a + 10*0 + 1*b

b0a ( in baza 10) = 100b + 10*0 + 1*a

adunam a0b si b0a, vom obtine:

100a + 10*0 + 1*b + 100b + 10*0 + 1*a

= 101a + 101b

= 101 (a+b)

-> 101 | 101(a+b)

pt ca 101 | 101 * x , unde x = a + b