Simple question for you lot to have a go at

[bloodhound]

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.

[alext87]

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

[blike]

0/99 = 0

[Dave]

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.

[bloodhound]

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.

[Dave]

That's probably the best way.

[wolfson]

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.

[Dave]

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.

 

[wolfson]

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