Sudhir's division by 12

If $p$ is a prime number greater than $12$ , it must leave a remainder of 1, 5, 7, or 11 when divided by 12. The remainder cannot be even, and it cannot be 3 or 9 since then the number would be divisible by 3. Whether $p\equiv1,5,7,11\pmod{12}$ , $p^2\equiv\boxed{1}\pmod{12}$ . This is because $1^2=1$ , $5^2=25=2\cdot 12+1$ , $7^2=49=4\cdot 12+1$ , and $11^2=121=10\cdot 12+1$ .

The remainder of a prime number modulo 12 can only be 1, 5, 7, or 11. If the remainder were even, the number would be even. If the remainder were divisible by 3, the number would be divisible by 3, so those are the only possible remainders.

$1^2 \equiv 1 \pmod{12}$

$5^2 \equiv 1 \pmod{12}$

$7^2 \equiv 1 \pmod{12}$

$11^2 \equiv 1 \pmod{12}$

So the answer is clearly $\fbox{1}$

There's a pattern. If you look at n^2(mod 12) where n is a prime greater than 3, you will find that n^2(mod 12) is always 1. Therefore, the answer is 1.

Let $p>3$ be a prime. Then it must be of the form $3n\pm 1$ . So we have...

$p^2=(3n\pm1)^2=9n^2\pm 6n +1\equiv 1 \pmod{3} ~~~~~~~~ \Longrightarrow ~~~~ 3\mid (p^2-1)$

We consider another form $4n\pm 1$ , and get...

$p^2=(4n\pm 1)^2=16n^2\pm8n+1\equiv 1 \pmod{4}~~~~~~ \Longrightarrow ~~~~ 4\mid (p^2-1)$

Hence, $12\mid (p^2-1)$ , implying...

$p^2\equiv\fbox{1}\pmod{12}$

any prime no. is of form ( 6n+1) or (6n-1) therefore, p^2 = 36 n^2 + 1 + 12n or 36 n^2 + 1 - 12n so dividing the above by 12 leaves a remainder 1 in all cases

This problem can be reduced using mods down numbers that are simpler to work with.
$12345 = 12000 + 240+ 60 + 36 + 9 mod 12 = 9 mod 12$
Now the question is: What is the remainder of $p^{2}$ after being divided by 12, given that p is the first prime after 9.
Obviously, the first prime after 9 is 11, so p=11.
Some basic math:
$p=11 mod 12$
$p^{2}= 121 mod 12$
$p^{2}= 1 mod 12$
So the final answer is $\boxed{1}$ since that was the final value of 121 mod 12.

the next prime number greater than 12345 is 12347 raised to power of 2 then divided by 12 the remainder is 1

The numbers after 12345 is 12346 but dividable perfectly by 2

When 12347 is squared and divided by 12, the answer will give ..... .08 , and I take 1 as answer

Let n be the remainder. Therefore, p^2 = n ( mod 12 ). Therefore, p^2 = n (mod 3 ) and p^2 = n (mod 4). Then, by quadratic residues, we know that all squares are congruent to 0 or 1 mod 4. Now, since p is a prime greater than 12345, it´s uneven, therefore it can´t be congruent to 0 mod 4, so it must be congruent to 1 mod 4. Therefore n = 1, so the residue is 1.

$12345 \equiv 9 \pmod {12}$ . A prime greater than $12345$ can be written as $12345 + 2k$ . $12345 + 2k \equiv {9 + 2k} \pmod {12}$ . $(12345 + 2k)^2 \equiv {(9 + 2k)^2} \pmod {12}$ . $9 + 2k$ can be written as $2x - 1$ . $(2x - 1)^2 = 4x^2 - 4x + 1$ . Subtract $1$ and you get $4x^2 - 4x = 4x(x-1)$ . In this point, we will have a remainder of $1$ unless you take an $x$ equal to $2 + 3m$ , for some integer $m \geq 0$ . This, because $\frac {4x(x - 1)} {12} = \frac {x(x-1)} {3}$ , and if $x$ or $x-1$ is divisible by $3$ , then $4x(x-1)$ is divisible by $12$ , and then the remainder of $\frac {4x^2 - 4x + 1} {12}$ is $1$ . We can affirm that $1$ is the answer, because we know the reminder will be $1$ because we can put in $x$ a lot of values, but not just that: you will always get a number ending, for example, in $7$ , and although we can't precise the exact prime, we know that there are primes greater than $12345$ ending with $7$ . Thus, the answer is $\boxed {1}$ .

I wish I explained correctly...

Step 1: You find the least prime no, greater than 12345 and that is 12373 Step 2: You square it -->(12373)^2 = 153091129 Step 3 : divide the result by 12 153091129, but first notice that 153091128/12 has no remainder, hence the remainder is 1.