I bet this is easy, but

if you(ie.me) don't know number theory , it's not.

here goes, is there an algebraic proof, so to speak, for this property?

56 + 9 = 65

I've figured out this much, but its not generalized.

10a + b + 9 = 10b + a...........where b = (a+1)

10a + (a+1) + 9

10a + a + 10

10(a+1) + a

10b + a = 10b + a

there is also the reverse where 10a + b -9 = 10b + a, if b = (a-1).

but how is it proven for any number of consecutive digits, when the number added or subtracted = 9*10^(k-2), where = k the number of digits

You need to visit The Church of the Flying Spaghetti Monster today.

Originally posted by eldragonfly
You need to visit The Church of the Flying Spaghetti Monster today.

Originally posted by joe shmo
and why should I do that?

Originally posted by eldragonfly
Bad joke. i don't understand what relationship you are trying to show, i can't see the connection that you are trying to show/imply.

Originally posted by joe shmo
oh,...so my math is not clear.....

if the digits are consecutive ....56, 45,34,67,89,23,ect....then adding 9 results in the reverse of the consecutive digits...so the first row plus nine = 65,54,43,76,98,32

hope that clears it up

>>how is it proven for any number of consecutive digits, when the number added or subtracted = 9*10^(k-2), where = k the number of digits

It does not hold in general, for k=3 for instance 123 + 90 = 213

Originally posted by apawn
>>how is it proven for any number of consecutive digits, when the number added or subtracted = 9*10^(k-2), where = k the number of digits

It does not hold in general, for k=3 for instance 123 + 90 = 213