Problems For National:1

[sourav das]

The concept of binary representation is great but i have some question(As I think those are bug or I couldn't understand)
i)"Notice that 2m||a-b if and only if the last m digits of a,b in binary representation are the same" I think it is not true. You should add that $m+1$th digits from right to left are different for fully divide part.
ii)for difference you need pairs. You find out how many numbers with different digits, do you count any pairs?
iii) Pairs of last 2 digit can be {(0,1),(1,1)}{(0,0),(1,0)} Similarly for others (last n digit) did you multiply these with $2^i$?

I started my solution saying it'll be nasty one. But i think my solution holds for the worst case of all $2n$. But the binary representation idea is nice one.

[FahimFerdous]

I don't think I made a typo.

Sorry, I was actually talking abt problem 5.

[FahimFerdous]

Yey, at last I solved problem 6, though Sourav helped me with a substitution.

[*Mahi*]

The concept of binary representation is great but i have some question(As I think those are bug or I couldn't understand)
i)"Notice that 2m||a-b if and only if the last m digits of a,b in binary representation are the same" I think it is not true. You should add that $m+1$th digits from right to left are different for fully divide part.
ii)for difference you need pairs. You find out how many numbers with different digits, do you count any pairs?
iii) Pairs of last 2 digit can be {(0,1),(1,1)}{(0,0),(1,0)} Similarly for others (last n digit) did you multiply these with $2^i$?

I started my solution saying it'll be nasty one. But i think my solution holds for the worst case of all $2n$. But the binary representation idea is nice one.

Thanks for reviewing my post carefully. (i) & (ii) were typos, and I edited them.
For (iii),
If there are $2^i$ binary numbers with different last $i$ digits, notice that for any number, there will be $2^{i-k-1}$ numbers only sharing the same last $k$ digits, as you can construct the last $k$ digits fixed, then a digit different, and then $2^{i-k-1}$ choices for the other digits.