Algebraic Proof question

Question

a, b, c are positive integers such that a>b>c

 

N is the largest three digit number that has the digits a, b and c.

K is the smallest three digit number that has the digits a, b and c.

 

(a) Use algebra to show that the difference between N and K is always a multiple of 99

 

(b) If a>b and b=c will the difference between N and K still be a multiple of 99?

Justify your answer.

 

Answer 1

987 and 789 are the numbers

N=abc=100a+10b+c

K=cba=100c+10b+a

N-K=100a+10b+c-100c-10b-a

N-K=99a-99c

N-K=99(a-c) which is a multiple of 99

 

988 and 889 are the numbers

N=abb=100a+10b+b

K=bba=100b+10b+a

N-K=100a+10b+b-100b-10b-a

N-K=99a-99b

N-K=99(a-b) which is a multiple of 99