# making $p$ the subject in $(3n + p) \bmod4 = 0$, how?

Let n and p be any positive integer, make $p$ the subject of the equation: $(3n + p)\bmod4 = 0$. How is it done?

I've worked out that the only values for p are 1, 2, 3 and 0.

This formula is for calculating the amount of padding required in a bitmap's pixel array:

Padding bytes (not necessarily 0) must be appended to the end of the rows in order to bring up the length of the rows to a multiple of four bytes. When the Pixel Array is loaded into memory, each row must begin at a memory address that is a multiple of 4. This address/offset restriction is mandatory only for Pixel Array's loaded in memory. For file storage purposes, only the size of each row must be a multiple of 4 bytes while the file offset can be arbitrary.[1] A 24-bit bitmap with Width=1, would have 3 bytes of data per row (blue, green, red) and 1 byte of padding, while Width=2 would have 2 bytes of padding, Width=3 would have 3 bytes of padding, and Width=4 would not have any padding at all.

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I don't understand the terminology: what does it mean to make a term that appears in an equation "the subject of the equation"? Do you mean you want to solve the congruence for $p$? If so, the answer is $p \equiv n \pmod 4$. – Pete L. Clark Apr 14 '11 at 16:37
Yes, you are absolutely correct, solve for congruence p. – meiryo Apr 14 '11 at 17:03

It depends on the value of $n$.

If $n \equiv 1$ (mod 4), then, you have $3+p\equiv0$ (mod 4) which means that $p\equiv1$ (mod 4). This procedure can be repeated for other values of $n$, namely $n\equiv0,1,2,3$ (mod 4)

In addition to this, note that if $p\equiv n$ (mod 4), then you have that $3n+p=4n\equiv0$ (mod 4)

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Thanks picakhu. – meiryo Apr 14 '11 at 17:02

$\rm\: mod\ 4\::\ \ 3\equiv -1\$ so $\rm\ 0\ \equiv\ p + 3\:n\ \equiv\ p - n\ \iff\ p\ \equiv\ n\ \$ by basic modular arithmetic.

Alternatively $\rm\ 4\ |\ p + 3\:n\ \iff\ 4\ |\ p - n + 4\:n\ \iff\ 4\ |\ p - n\ \$ where $\rm\ a\ |\ b\$ means $\rm\:a\:$ divides $\rm\:b\:.$

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