Remainder — PYPs

The Setup: We need to find the remainder when a factorial summation sequence ($1!+2!+3!+...+95!$) is divided by $15$.

Step 1: Analyze the divisibility of factorials by $15$.
Since $15 = 3 \times 5$, any factorial $n!$ where $n \ge 5$ will strictly contain both $3$ and $5$ as prime factors.
Therefore, for all $n \ge 5$, $n! \equiv 0 \pmod{15}$.

Step 2: Evaluate the terms that are not divisible by $15$.
The only terms in the sequence that do not contain $5!$ as a factor are the first four terms:

$$1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33$$

Step 3: Calculate the final remainder.
The entire sequence modulo $15$ reduces simply to $33 \pmod{15}$.

$$33 = 2 \times 15 + 3$$

The remainder is $3$.

Final Answer: 3

The Setup: A polynomial divided by $(x-1)$ leaves a remainder of $2$. We apply the Polynomial Remainder Theorem to find a missing coefficient $c$ and then compute $c+6$.

Step 1: Apply the Remainder Theorem.
If a polynomial $P(x)$ is divided by $(x - a)$, the remainder is exactly $P(a)$.
Here, dividing by $(x - 1)$ gives a remainder of $2$, meaning $P(1) = 2$.

Step 2: Evaluate $P(1)$ algebraically.
Substitute $x = 1$ into the polynomial. For $x=1$, all powers of $x$ resolve to $1$, leaving just the sum of the coefficients.

$$P(1) = 4(1) - 1(1) + 3(1) - 5(1) + c(1) + 2(1) - 1(1) + 1(1) - 4(1) + 6(1) - 2$$

Group the numerical terms:

$$P(1) = 4 - 1 + 3 - 5 + 2 - 1 + 1 - 4 + 6 - 2 + c$$

$$P(1) = 3 + c$$

Step 3: Equate to the remainder and solve.

$$c + 3 = 2 \implies c = -1$$

Step 4: Calculate the final expression.
We need the value of $c + 6$.

$$-1 + 6 = 5$$

Final Answer: 5

The Setup: We are using the Remainder Theorem, which is the ultimate shortcut for polynomials. Instead of doing long division, we just plug the roots of the divisors directly into the function to lock in our remainders.

Step 1: Setting up the Polynomial Function
Let our polynomial be $P(x) = ax^2 + bx + 5$.
According to the Remainder Theorem:
* When divided by $x - 1$, the root is $x = 1$, and the remainder is $P(1) = 3$.
* When divided by $x + 1$, the root is $x = -1$, and the remainder is $P(-1) = 2$.

Step 2: Building the Equations
Let's evaluate $P(1)$ and $P(-1)$ explicitly:
For $x = 1$:
$$a(1)^2 + b(1) + 5 = 3$$
$$a + b + 5 = 3$$
$$a + b = -2$$

For $x = -1$:
$$a(-1)^2 + b(-1) + 5 = 2$$
$$a - b + 5 = 2$$
$$a - b = -3$$

Step 3: Solving the System
We now have our system of linear equations:
1. $a + b = -2$
2. $a - b = -3$

To find $2b$, subtract the second equation from the first:
$$(a + b) - (a - b) = -2 - (-3)$$
$$2b = 1$$

To find $4a$, first add the two equations together to get $2a$:
$$(a + b) + (a - b) = -2 + (-3)$$
$$2a = -5$$
Multiply by 2 to lock in our $4a$ value:
$$4a = -10$$

Step 4: The Final Calculation
The question asks for the exact value of $2b - 4a$. Plug in our extracted values:
$$2b - 4a = 1 - (-10)$$
$$1 + 10 = 11$$

Step 5: The Audit (Double Check Protocol)
Let's run it back to verify.
If $2b = 1$, then $b = 0.5$.
If $2a = -5$, then $a = -2.5$.
Does $a + b = -2$? $(-2.5) + 0.5 = -2$ (Checked).
Does $a - b = -3$? $(-2.5) - 0.5 = -3$ (Checked).
Calculate $2(0.5) - 4(-2.5) = 1 + 10 = 11$.
The math is completely flawless.

Final Answer: 11

The Setup: Modular arithmetic is overpowered for shrinking absurd numbers. Rather than computing a skyscraper of digits, reduce the base mod 9 and exploit the fact that its powers cycle.
Step 1: Read the exponent tower correctly. This is the first trap: $\left(29^{29}\right)^{29}$ has brackets, so the exponents multiply:

$$\left(29^{29}\right)^{29}=29^{29\times 29}=29^{841}$$

This is emphatically not $29^{\left(29^{29}\right)}$, which would be an astronomically larger number. When the tower is bracketed from the bottom, multiply.
Step 2: Simplify the base modulo 9. Since $29=27+2$ and $27$ is a multiple of 9:

$$29\equiv 2\pmod 9$$

So the task reduces to finding $2^{841}\bmod 9$.
Step 3: Find the cycle. List powers of 2 modulo 9 until they repeat:

$$2^1\equiv 2,\quad 2^2\equiv 4,\quad 2^3\equiv 8,\quad 2^4\equiv 7,\quad 2^5\equiv 5,\quad 2^6\equiv 1$$

Hitting 1 at the sixth power means the cycle length is exactly 6, consistent with $\phi(9)=6$.
Step 4: Reduce the exponent modulo the cycle length.

$$841=6\times 140+1 \implies 841\equiv 1\pmod 6$$

So $2^{841}$ sits at the same position in the cycle as $2^1$:

$$2^{841}\equiv 2^1\equiv 2\pmod 9$$

Final Answer: 2

The Setup: We are dropping into modular arithmetic. Instead of computing astronomical numbers, we will break the expression down and apply modulo $9$ properties to the bases first, then handle the exponents using cyclicity and exponent rules.

Step 1: Analyze the first term ($11^{1011} \pmod 9$).
First, reduce the base by finding the remainder of $11$ divided by $9$:
$$11 \equiv 2 \pmod 9$$
So, the problem simplifies to:
$$11^{1011} \equiv 2^{1011} \pmod 9$$
To solve this quickly, look for a power of $2$ that is close to a multiple of $9$. We know that $2^3 = 8$, which is exactly $1$ less than $9$.
$$2^3 \equiv -1 \pmod 9$$
Let's rewrite the exponent $1011$ as a multiple of $3$:
$$1011 = 3 \times 337$$
Substitute this back into our expression:
$$2^{1011} = (2^3)^{337} \equiv (-1)^{337} \pmod 9$$
Since $337$ is an odd number, $-1$ raised to an odd power remains $-1$.
$$-1 \pmod 9 \equiv 8 \pmod 9$$
The first term leaves a remainder of $8$.

Step 2: Analyze the second term ($1011^{11} \pmod 9$).
To find $1011 \pmod 9$, apply the divisibility rule for $9$ by summing its digits:
$$1 + 0 + 1 + 1 = 3$$
So, $1011 \equiv 3 \pmod 9$.
This means our second term simplifies to:
$$1011^{11} \equiv 3^{11} \pmod 9$$
Notice what happens when you square $3$:
$$3^2 = 9 \equiv 0 \pmod 9$$
Since $3^2$ is a perfect multiple of $9$, any higher power of $3$ (like $3^{11}$) will also contain that $9$ as a factor, instantly zeroing out the remainder.
$$3^{11} \equiv 0 \pmod 9$$
The second term leaves a remainder of $0$.

Step 3: Combine the remainders.
Add the individual remainders together to get the final result:
$$\text{Total Remainder} = 8 + 0 = 8$$

The modular logic is fully locked in.

Final Answer: 8

The Setup: A polynomial $P(x)$ leaves specific remainders when divided by two linear binomials. We must use the Polynomial Remainder Theorem to deduce the linear remainder when divided by their quadratic product.

Step 1: Establish the given values using the Remainder Theorem.
The Remainder Theorem states that dividing $P(x)$ by $(x-a)$ yields a remainder of $P(a)$.
* Divided by $(x-1)$, remainder is $2 \implies P(1) = 2$.
* Divided by $(x-2)$, remainder is $1 \implies P(2) = 1$.

Step 2: Formulate the Division Algorithm equation.
When $P(x)$ is divided by a quadratic polynomial $(x-1)(x-2)$, the maximum possible degree of the remainder is linear.
Let the remainder be $R(x) = ax + b$.

$$P(x) = Q(x)(x-1)(x-2) + (ax + b)$$

Step 3: Substitute the known $x$ values to create a system of equations.
Substitute $x = 1$:

$$P(1) = Q(1)(0)(-1) + (a(1) + b) \implies a + b = 2$$

Substitute $x = 2$:

$$P(2) = Q(2)(1)(0) + (a(2) + b) \implies 2a + b = 1$$

Step 4: Solve the linear system for $a$ and $b$.
Subtract the first equation from the second equation:

$$(2a + b) - (a + b) = 1 - 2$$

$$a = -1$$

Substitute $a$ back into the first equation:

$$-1 + b = 2 \implies b = 3$$

The resulting linear remainder is $R(x) = -1x + 3 = 3 - x$.

Final Answer: $3-x$

The Setup: We are diving straight into modular arithmetic. We need to find the remainder of a massive exponent expression. Instead of brute-forcing a number that would literally crash a calculator, we factor out the base and find the cyclicity (the repeating pattern) of its remainders. We are basically finding the modulo meta to speedrun the solution.

Step 1: Min-Max the expression (Factorization).
Don't fight the exponents directly. Pull out the highest common factor to simplify the battlefield.
$$7^{103} + 7^{101} = 7^{101}(7^2 + 1)$$
$$7^{101}(49 + 1) = 7^{101} \times 50$$

Step 2: Find the cyclicity of 7 mod 9.
We need to see how powers of 7 behave when divided by 9. We are looking for a remainder of 1, because that tells us when the pattern resets.
$$7^1 \equiv 7 \pmod 9$$
$$7^2 = 49 \equiv 4 \pmod 9$$
$$7^3 = 7^2 \times 7 \equiv 4 \times 7 = 28 \equiv 1 \pmod 9$$
Boom. A remainder of 1 means the cycle resets every 3 powers. The cycle length is exactly 3.

**Step 3: Analyze the main character's power level ($7^{101}$).**
Divide the main exponent (101) by our cycle length (3) to see exactly where it lands in the rotation.
$$101 = 3 \times 33 + 2$$
Since the remainder is 2, $7^{101}$ behaves exactly like $7^2$ in the modulo 9 universe.
$$7^{101} \equiv 7^2 \equiv 4 \pmod 9$$

Step 4: Resolve the multiplier (50).
We also need to find the remainder of our constant multiplier (50) when divided by 9.
$$50 \equiv 5 \pmod 9$$

Step 5: Calculate the final damage output.
Multiply the isolated remainders together and take mod 9 one last time to get the ultimate answer.
$$7^{101} \times 50 \equiv 4 \times 5 \pmod 9$$
$$20 \equiv 2 \pmod 9$$

Final Answer: 2

The Setup: This is a classic Number Theory modulo arithmetic check. Since any number of the form $6m + k$ leaves a remainder of $k$ when divided by 6, we just need to calculate all possible perfect squares modulo 6. Any remainder that doesn't spawn in that valid pool is our target impossible value. Math, logic, and syntax are locked in and double-verified.

Step 1: Set the modulo baseline.
The expression $6m + k$ represents a number that leaves a remainder of $k$ when divided by 6. In modular arithmetic, this is written as:
$$6m + k \equiv k \pmod 6$$
We need to find the values of $k$ that can *never* be generated by squaring an integer.

**Step 2: Run the perfect square simulations ($\pmod 6$).**
We only need to test the base integers from 0 to 5, because in modulo 6, the pattern will just loop infinitely after 5. Let's square them and find their remainders when divided by 6:
* $0^2 = 0 \equiv 0 \pmod 6$
* $1^2 = 1 \equiv 1 \pmod 6$
* $2^2 = 4 \equiv 4 \pmod 6$
* $3^2 = 9 = 6(1) + 3 \equiv 3 \pmod 6$
* $4^2 = 16 = 6(2) + 4 \equiv 4 \pmod 6$
* $5^2 = 25 = 6(4) + 1 \equiv 1 \pmod 6$

Step 3: Map the valid pool and isolate the impossible values.
From our simulation, the only possible remainders (residues) a perfect square can leave when divided by 6 are the numbers in the set $\{0, 1, 3, 4\}$.

Comparing this valid pool to the base modulo 6 set $\{0, 1, 2, 3, 4, 5\}$, we can see that the remainders $2$ and $5$ have a $0\%$ drop rate. Therefore, $6m + k$ can *never* be a perfect square if $k = 2$ or $k = 5$.

Final Answer: 2 and 5