# I bet this is easy, but

joe shmo
Posers and Puzzles 27 Sep '08 18:10
1. joe shmo
Strange Egg
27 Sep '08 18:101 edit
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
2. eldragonfly
leperchaun messiah
27 Sep '08 20:44
You need to visit The Church of the Flying Spaghetti Monster today.
3. joe shmo
Strange Egg
27 Sep '08 21:21
Originally posted by eldragonfly
You need to visit The Church of the Flying Spaghetti Monster today.
and why should I do that?
4. eldragonfly
leperchaun messiah
27 Sep '08 21:351 edit
Originally posted by joe shmo
and why should I do that?
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.
5. joe shmo
Strange Egg
27 Sep '08 21:39
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.
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
6. wolfgang59
29 Sep '08 11:40
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
you are trying to prove that

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

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

ðŸ™„
7. 01 Oct '08 02:27
>>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
8. joe shmo
Strange Egg
01 Oct '08 02:41
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
ahh,..I guess if it did 123+90 would have to = 321

ok, Thanks