a)
( 1 a + a 1 + 1 b + b1) l ( a + b + 2)
= 10 + a + 10 a + 1 + 10 + b + 10 b + 1 =
= 11 a + 11 b + 22 =
= 11 ( a + b + 2) l ( a + b + 2)
b)
( abc - bc ) l 100
= 100 a + 10 b + c - ( 10 b + c ) =
= 100 a + 10 b - 10 b + c - c =
= 100 a l 100 → cctd ( ceea ce trebuia demonstrat)
c)
b l ( 1 ab - 10 · 1 a )
( 1 ab - 10 · 1 a ) = 100 + 10 a + b - 10 × ( 10 + a) =
= 100 + 10 a + b - 100 - 10 a =
= b l b ( b divizibil cu b )
d)
b > c , atunci 9 l ( abc - acb )
abc - acb = 100 a + 10 b + c - ( 100 a + 10 c + b ) =
= 100 a - 100 a + 10 b - b + c - 10 c =
= 9 b - 9 c =
= 9 ( b - c ) ⇒ b > c l 9
