Urgend: Sum of digits question

[Mathman23]

I have question here which has puzelled me since monday. Hope there is somebody here who can give a hint/help.

Let "a" be number written in base 10.

a_0 * 10^0 + a_1 * 10^1 + a_2 * 10^2 + -------+ a_n * 10^n

where 0 \leq a_i \leq 10.

Prove that the number 2 divides a, if and only if 2 divides a_0.

Can anybody give me a hint on how to go about solving this?

Can I claim that a_0 must not equal zero for 2 to divide a_0 ??

Sincerley Yours
Fred

 

[matt grime]

2 divides 0, 10, 20, 100,000, all of which have a_0=0.

x divides y if and only if y is congruent to zero mod x.

All its saying is that a number is even if its units entry is even, which you've known since you were so high (holds hand some height above floor).

 

[Mathman23]

Maybe I'm stupid then it comes to this kind of math, but what should my approach be then I'm faced with such an assigment as this? Because my textbook doens't explain all this very well.

What would You Matt surgest then solving this kind of problem?

What I get in this problem is that the divider must not be larger than 10, or smaller than 0.

Sincerely

Fred

 

[matt grime]

No. I don't think you understand the question, do you? Try an example.

Suppose a=423526, then what are the a_i? They are a_0=6, a_1=2, a_3=5, a_4=3, a_5=2, a_6=4.

Now, you've known all your mathematical life since you learned what division was that a is divisible by 2 precisely because a_0=6 (i.e. the last digit in the decimal expansion) is divisible by 2. a = a_0 +10*(something) and 10 is divisible by 2.

 

[Mathman23]

Okay thanks for the example,

Than You prove this by saying that as long a the rightest digit in the binary number a_0 is dividable by 2, then the whole of a is dividable by 2??

Sincerley
Fred

p.s. In other words as longs as 2 is dividable by a number y with the smallest reminder, then 2 is dividable by the whole number "a"?

 

[matt grime]

This is not binary, this is base 10.

Dividable is not, I think, a word. Divisible is, and you have it the wrong way round. You are asking when a is divisible by 2.

 

[HallsofIvy]

Well, you don't "prove" it that way. Write your number
a_na_n-1...a₂a₁a₀
as (a_na_n-1...a₂a₁)(10)+ a₀.

Do you see why that's true? Do you see why the first term is always divisible by 2?