who wants to help Scorch get a better grade? :)

[SWScorch]

yeah yeah, I know, you don't do people's homework here. But what about extra credit, eh? I have this question for 15 extra points on my latest discrete test, and let me tell you, I need it. Trouble is, I'm not sure where to start. Here's la pregunta:

show that any positive integer is divisible by three if and only if the sum of its decimal digits is divisible by three. Use any positive integer is equal to a(n)a(n-1)...a(2)a(1)a(0) and 10^k is congruent to 1(mod 3) for k = 1,2,3...

Okay, what the hell is he talking about here? Granted, I suck at proofs regardless, but this is a doozie. Anyone wanna at least get me started? Pleeease?

 

[Jassi]

I would ask my grade school teacher for you but she is about 10k miles away. I guess I always took that for granted, if you get a proof, share

 

[jst0ney]

too hard for a weekend.

 

[Falloutboy]

umm no I hate math to begin with and I've got too much booze in me right now even if I did want to help

 

[jspeicher]

you are so not getting any help tonight. :brokenheart:

 

[Injury]

If the sum of the numbers in any integer is divisible by 3, then the number itself will be, too... but I can't read past the first sentence in the third paragraph without thinking, so no way in hell I'm gonna try to figure out what it is saying.

 

[aux]

divisible by 3 is the same as congruent to 0 (mod 3)

a(n)a(n-1)...a(2)a(1)a(0) = a(n)*10^n + a(n-1)*10^(n-1) + ... + a(2)*10^2 + a(1)*10^1 + a(0)*10^0
which is congruent to a(n)*1 + a(n-1)*1 + ... + a(2)*1 + a(1)*1 + a(0)*1 (mod 3)
because 10^k is congruent to 1 (mod 3) as given

therefore a(n)a(n-1)...a(2)a(1)a(0) is divisible by 3, or the same, congruent to 0 (mod 3) if and only if
a(n)*1 + a(n-1)*1 + ... + a(2)*1 + a(1)*1 + a(0)*1 is congruent to 0 (mod 3), which is the same as a(n)*1 + a(n-1) + ... + a(2)*1 + a(1)*1 + a(0)*1 divisible by 3

a(n)*1 + a(n-1)*1 + ... + a(2)*1 + a(1)*1 + a(0)*1 is the sum of the digits

edit: 10^k is congruent to 1 (mod 3) also for k = 0, because 10^0 = 1; as given 10^k is congruent to 1 (mod 3) also for k = 1, 2, ...

 

[Falloutboy]

Originally posted by: aux
divisible by 3 is the same as congruent to 0 (mod 3)

a(n)a(n-1)...a(2)a(1)a(0) = a(n)*10^n + a(n-1)*10^(n-1) + ... + a(2)*10^2 + a(1)*10^1 + a(0)*10^0
which is congruent to a(n)*1 + a(n-1)*1 + ... + a(2)*1 + a(1)*1 + a(0)*1 (mod 3)
because 10^k is congruent to 1 (mod 3) as given

therefore a(n)a(n-1)...a(2)a(1)a(0) is divisible by 3, or the same, congruent to 0 (mod 3) if and only if
a(n)*1 + a(n-1)*1 + ... + a(2)*1 + a(1)*1 + a(0)*1 is congruent to 0 (mod 3), which is the same as a(n)*1 + a(n-1) + ... + a(2)*1 + a(1)*1 + a(0)*1 divisible by 3

a(n)*1 + a(n-1)*1 + ... + a(2)*1 + a(1)*1 + a(0)*1 is the sum of the digits

edit: 10^k is congruent to 1 (mod 3) also for k = 0, because 10^0 = 1; as given 10^k is congruent to 1 (mod 3) also for k = 1, 2, ...

am I supposed to understand what this means?? also what level math is this (just checking to see how much I've forgotten)

 

[SWScorch]

I'm taking Discrete 1. aux, thanks, I'll see if I can work through that. This stuff confuses me so much.