Why b%2==1 implies that rightmost digit of b in binary form is 1 How one can deduce that if b is any number and if b%2==1 then rightmost digit of b in binary form will be 1 without checking it manualy
 A: Expressing $b$ in binary we get $$b=b_n2^n+b_{n-1}2^{n-1}+\cdots+b_22^2+b_12^1+b_02^0$$
where each of $b_n, b_{n-1},\ldots, b_1$ are either $0$ or $1$.  Now, when we take this modulo $2$, every summand disappears except for $b_02^0=b_0$, which is the rightmost binary digit of $b$.
A: HINT :
Let $$b=b_0+2b_1+2^2b_2+...+2^nb_n$$
What do you get when you calculate $b\bmod 2$?
A: b % 2 means "the remainder when you divide by 2".
A binary number like $101001$ can be written:
$$1 * 2^5 + 0 * 2^4 + 1 * 2^3 + 0 * 2^2 + 0 * 2^1 + 1*2^0$$
which is
$$1 * 32 + 0 * 16 + 1 * 8 + 0 * 4 + 0 * 2 + 1*1$$
If you get rid of everything that is divisible by $2$ (since it will all go away when you divide by $2$ and take the remainder) you get:
$$1*1 = 1$$
If the last digit of our number was $0$, we would have gotten zero instead.
This is the same as saying a number is even if and only if its least significant bit is 0 (binary) -- otherwise it's odd.
A: You only need to demonstrate that it is 1 when there is only 1 digit to prove that it will be 1 for any odd number, regardless of the number of digits.
It's a base two number: the last digit can only be 1 or 0. When you add 1 to 0 it will go to 1 and when you add another it goes back to 0 and the number of digits increases by 1.  Since it can only be one or the other, and zero is even, then all odd numbers will have a 1 as the right most digit.
