Take a four digit positive number A, reverse the digits to make number B

 

show that A^2 - B^2 is always divisible by 99

 

example: A=3785 B=5873

 

first question in my analysis exam today , dont know why.

Does not work. Take A=1001 then B=1001 therefore Ans= 0 which is not divisble by 99.

0/99 = 0

Write a = 10^3*x₁ + 10^2*x₂ + 10*x₃ + x₄ and b similarly, expand the expression and show the coefficients are all factors of 99 is the first method that comes to mind.

dave, thats the first thing that came into my mind as well, but it takes hell of a job to expand A2-B2

 

the way i did it was to show that A-B is divisible by 9 and A+B is divisible by 11

 

i wonder if there are any better ways to work round it. i was about to use modular arithmetic but i am rubbish at that.

That's probably the best way.

Try writing A in the form

 

A=1000a+100b+10c+d,

 

where a,b,c and d are the digits of A's decimal expansion. Then see what happens if you reverse and subtract.

 

Ask if you need a further hint.

Try writing A in the form

 

A=1000a+100b+10c+d' date='[/quote']

 

Write a = 10^3*x1 + 10^2*x2 + 10*x3 + x4 and b similarly, expand the expression and show the coefficients are all factors of 99 is the first method that comes to mind.

 

Opps didnt see it i appologise. Should'nt just read question. lol