Proof question (1 Viewer)

[Run hard@thehsc]

How would you do and approach this question?

 

[idkkdi]

Suppose 10^k ak + 10^k-1ak-1 +... + a0 = 3m, mEZ
10^k ak + 10^k-1ak + ... + a0 - (ak + ak-1 + ... + a0) = 3m - (ak + ak-1+....+a0)
9(................) using the x^n-y^n factorisation = 3m - (ak+ak-1+...+a0)
Since LHS is divisible by 3, RHS is, so ak+ak-1+...+a0 is divisible by 3.

Alternatively, suppose
ak + ak-1 + ... + a0 = 3p, mEZ
-(ak + ak-1 + ... + a0) = -3p
(10^k ak + 10^k-1ak + ... + a0) - (ak + ak-1 + ... + a0) = (10^k ak + 10^k-1ak + ... + a0) -3p
Same reasoning as above follows.

 

[5uckerberg]

Suppose 10^k ak + 10^k-1ak-1 +... + a0 = 3m, mEZ
10^k ak + 10^k-1ak + ... + a0 - (ak + ak-1 + ... + a0) = 3m - (ak + ak-1+....+a0)
9(................) using the x^n-y^n factorisation = 3m - (ak+ak-1+...+a0)
Since LHS is divisible by 3, RHS is, so ak+ak-1+...+a0 is divisible by 3.

Alternatively, suppose
ak + ak-1 + ... + a0 = 3p, mEZ
-(ak + ak-1 + ... + a0) = -3p
(10^k ak + 10^k-1ak + ... + a0) - (ak + ak-1 + ... + a0) = (10^k ak + 10^k-1ak + ... + a0) -3p
Same reasoning as above follows.

Note instead here you can state that 9(................) = where has to be divisible by 9 and there it is obvious that it is also divisible by 3 and solved. ak+ak-1+...+a0 is divisible by 3. To me this statement becomes redundant when something is divisible by 9. SOmetimes you can probably get away with such things.

From a blink of an eye, this has to be Matrix homework.