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1 mod 3 equals 1, since 1/3 = 0 with a remainder of 1. To find 1 mod 3 using the modulus method, we first find the highest multiple of the divisor, 3 that is equal to or less than the dividend, 1. Then, we subtract the highest multiple from the dividend to get the answer to 1 mod 3. Multiples of 3 are 0, 3, 6, 9, etc. and the highest multiple ...
This is more common in math than in e.g. computer science but it is worth keeping in mind because it simplifies a lot. This is not to say that the choice of 0, 1, 2, ..., b-1 as representatives of a value of a mod b (when a is an integer and "a mod b" is a function output) is not a natural one, which may be what is going on here.
There seems to be a fundamental difference between $2\pmod3$ and $1\pmod3$ in this way (similarly, between $3\pmod4$ and $1\pmod4$). $\endgroup$ – Greg Martin Commented Apr 28, 2017 at 20:07
We can check that $0^2\cong0\pmod 3$, $1^2\cong1\pmod 3$, and $2^2\cong4\cong1 \pmod3$. There are only three cases to check. There are only three cases to check. Share
Let n be even, then n2n + 1 ≡ n(1) + 1 (mod3) ≡ n + 1 (mod 3) We want this to be 0 mod3, so n ≡ 2 (mod3). We have the following conditions on n, n ≡ 0 (mod 2) and n ≡ 2 (mod 3). This can be consolidated as n ≡ 2 (mod 6). Likewise you can deal with the other case. This was very helpful, thanks.
Let's call the set of integers modulo 3 by $\mathbb{F}_3.$ It has three elements, which we will call $\{\overline{0},\overline{1},\overline{2}\}.$ Don't confuse these with $0, 1,2\in\mathbb{Z},$ as they're quite different! One way to think of them is that each one represents the set of all integers which has the given remainder when divided by 3.
$23\mod 24\equiv -1\mod 24$ considering squares, we have . $1^2\mod 24\equiv 1\mod 24$ $5^2\mod 24\equiv 1\mod 24$ $7^2\mod 24\equiv 1\mod 24$ $11^2\mod 24\equiv 1\mod 24$ For same reason as above, squares of below congruence classes are equal to $1\mod 24$. $13\mod 24\equiv -11\mod 24$ $17\mod 24\equiv -7\mod 24$ $19\mod 24\equiv -19\mod 24$
Using the hint is to try the three cases: Case 1: $n \equiv 0 \mod 3$ Remember if $a \equiv b \mod n$ then $a^m \equiv b^m \mod n$ [$*$]
Suppose $p = 1 \bmod 3$, prove the following statements: prove that $x^2 + x + 1 = 0 \mod p$ has a solution ...
That's just a fancy way of writing “ $2$ raised to the power of something even will be $1$ mod $3$ and when raised to an odd power, it will be $2$ mod $3$ ” Now, $2^{2006}$ is a very even number (make sure you understand why). Therefore, the answer to your question is: $$ 2^{2^{2006}}=2^{even}\equiv 1\ (\text{mod}\ 3) $$ PS: You did manage ...