- 27 Sep '08 18:10 / 1 editif 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 - 27 Sep '08 21:39

oh,...so my math is not clear.....*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.**

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 - 29 Sep '08 11:40

you are trying to prove that*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

{10a + (a+1)} + 9 = {10(a+1) + a}

since both sides of the equation are 11a + 1 its QED

- 01 Oct '08 02:41

ahh,..I guess if it did 123+90 would have to = 321*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

ok, Thanks