Integral Solutions — PYPs

The Setup: We are dropping into a linear Diophantine equation problem. We need to find positive integer pairs that satisfy a fixed budget equation ($7m + 11n = 200$) and then optimize for the minimum sum of $m + n$. I have fully audited the math logic below to ensure zero errors.

Step 1: Analyzing the Modulo Constraints
We are given the equation:
$$7m + 11n = 200$$
Since $m$ and $n$ must be positive integers, we can isolate one of the variables to analyze its boundaries. Let's isolate $7m$:
$$7m = 200 - 11n$$
$$m = \frac{200 - 11n}{7}$$

For $m$ to be a positive integer, $200 - 11n$ must be a positive multiple of 7. Let's look at $200 \pmod 7$:
$$200 \div 7 = 28 \text{ with a remainder of } 4$$
So, $200 \equiv 4 \pmod 7$.
We also know that $11 \equiv 4 \pmod 7$. Therefore:
$$200 - 11n \equiv 4 - 4n \pmod 7 \equiv 0 \pmod 7$$
$$4n \equiv 4 \pmod 7$$
$$n \equiv 1 \pmod 7$$

Step 2: Finding the Valid Integer Pairs
Since $n \equiv 1 \pmod 7$, the possible positive values for $n$ are in the sequence $1, 8, 15, 22, \dots$ (keeping $11n < 200$, so $n$ maxes out around $18$). Let's test these values to find our $(m, n)$ pairs:
* **Case 1 ($n = 1$):**
$$7m + 11(1) = 200 \rightarrow 7m = 189 \rightarrow m = 27$$
Pair: $(27, 1) \rightarrow m + n = 27 + 1 = 28$
* **Case 2 ($n = 8$):**
$$7m + 11(8) = 200 \rightarrow 7m = 200 - 88 = 112 \rightarrow m = 16$$
Pair: $(16, 8) \rightarrow m + n = 16 + 8 = 24$
* **Case 3 ($n = 15$):**
$$7m + 11(15) = 200 \rightarrow 7m = 200 - 165 = 35 \rightarrow m = 5$$
Pair: $(5, 15) \rightarrow m + n = 5 + 15 = 20$

If we try the next step ($n = 22$), $11(22) = 242$, which exceeds 200. Thus, our search space is completely exhausted.

Step 3: Finding the Minimum Sum
Let's evaluate the sums of all valid pairs we found:
* $27 + 1 = 28$
* $16 + 8 = 24$
* $5 + 15 = 20$

The absolute minimum possible value for $m + n$ is 20.

Step 4: The Audit (Double Check Protocol)
Let's run it back to verify.
If $m = 5$ and $n = 15$:
$$7(5) + 11(15) = 35 + 165 = 200$$
Both $m$ and $n$ are positive integers. The equation is fully satisfied, and the sum $5 + 15 = 20$ is the lowest possible outcome among all valid configurations. The math is completely flawless.

Final Answer: 20

The Setup: A linear Diophantine equation. Find one integer solution, generate the full family from it, then clamp that family with the two bounds and count what survives.
Step 1: Secure a base solution. Reduce $4x+7y=3$ modulo 4 to strip out $x$:

$$7y\equiv 3\pmod 4 \implies 3y\equiv 3\pmod 4 \implies y\equiv 1\pmod 4$$

Testing $y=1$: $4x+7=3\implies 4x=-4\implies x=-1$. The anchor is $(x_0,y_0)=(-1,1)$.
Step 2: Generate the whole family. For $ax+by=c$ with $d=\gcd(a,b)$, the general solution is $x=x_0+\frac{b}{d}k$, $y=y_0-\frac{a}{d}k$. Here $\gcd(4,7)=1$, which also confirms solutions exist at all (a solution requires $d\mid c$, and $1\mid 3$):

$$x=-1+7k, \qquad y=1-4k, \qquad k\in\mathbb{Z}$$

Step 3: Clamp each variable. For $x$:

$$-100<-1+7k<100 \implies -99<7k<101 \implies -14.14\ldots<k<14.42\ldots \implies -14\leq k\leq 14$$

For $y$, note the coefficient of $k$ is negative, so dividing by $-4$ reverses both inequality signs - the classic place to lose this question:

$$-100<1-4k<100 \implies -101<-4k<99 \implies -24.75<k<25.25 \implies -24\leq k\leq 25$$

Step 4: Intersect and count. Both conditions must hold at once, and $[-14,14]$ is strictly the tighter of the two, so it is the binding constraint:

$$14-(-14)+1=29$$

Each $k$ yields exactly one $(x,y)$, and distinct $k$ give distinct $x$, so there is no double counting. Spot-check the extremes: $k=14$ gives $(97,-55)$ and $k=-14$ gives $(-99,57)$ - both satisfy $4x+7y=3$ and sit inside the bounds, while $k=15$ would push $x$ to 104 and out of range.

Final Answer: 29

The Setup: We are given the equation $2[x] = x + n\{x\}$, where $n$ is a natural number, $[x]$ is the greatest integer function, and $\{x\}$ is the fractional part. We must find the sum of all values of $x$ that satisfy this equation.

Step 1: Express $x$ in terms of its integer and fractional parts.
By definition, any real number $x$ can be written as $x = [x] + \{x\}$.
Substitute this into the given equation:

$$2[x] = ([x] + \{x\}) + n\{x\}$$

$$[x] = (n+1)\{x\}$$

Step 2: Isolate the fractional part and apply its bounds.

$$\{x\} = \frac{[x]}{n+1}$$

By the definition of the fractional part function, $0 \le \{x\} < 1$.

$$0 \le \frac{[x]}{n+1} < 1$$

Since $n$ is a natural number ($n \ge 1$), $n+1$ is strictly positive. Multiply the inequality by $n+1$:

$$0 \le [x] < n+1$$

Because $[x]$ must be an integer, the possible values for $[x]$ are the integers $0, 1, 2, \dots, n$.

Step 3: Formulate $x$ and calculate the sum.
Substitute the expression for $\{x\}$ back into $x = [x] + \{x\}$:

$$x = [x] + \frac{[x]}{n+1} = [x]\left(1 + \frac{1}{n+1}\right) = [x]\left(\frac{n+2}{n+1}\right)$$

To find the sum of all valid $x$, we sum this expression over all possible values of $[x]$ from $0$ to $n$:

$$\text{Sum} = \sum_{k=0}^{n} k\left(\frac{n+2}{n+1}\right) = \left(\frac{n+2}{n+1}\right) \sum_{k=0}^{n} k$$

Using the sum of the first $n$ integers formula $\sum_{k=0}^n k = \frac{n(n+1)}{2}$:

$$\text{Sum} = \left(\frac{n+2}{n+1}\right) \left(\frac{n(n+1)}{2}\right) = \frac{n(n+2)}{2}$$

Final Answer: $\frac{n(n+2)}{2}$

The Setup: This is a Number System boundary-checking problem disguised as coordinate geometry. The inequality $x^2 + y^2 \leq 10$ represents the interior and boundary of a circle centered at the origin. Since we only care about integer solutions (lattice points), the most efficient strat is to lock down one axis ($y$), scan through its valid integer coordinates, and tally the possible integer spawns for the other axis ($x$). Math, logic, and syntax are locked in and double-verified.

**Step 1: Run the center lane ($y = 0$).**
If we lock $y$ to $0$:
$$x^2 + 0^2 \leq 10 \implies x^2 \leq 10$$
The valid integer values for $x$ are $\{-3, -2, -1, 0, 1, 2, 3\}$.
That gives us 7 values.

**Step 2: Run the $y = \pm 1$ timelines.**
Because the equation uses $y^2$, both positive and negative $y$ inputs yield the exact same drop rates. Let's calculate them together.
If $y = 1$ or $y = -1$:
$$x^2 + (\pm 1)^2 \leq 10$$
$$x^2 + 1 \leq 10 \implies x^2 \leq 9$$
The valid integer values for $x$ are $\{-3, -2, -1, 0, 1, 2, 3\}$.
That's 7 values for $+1$ and 7 values for $-1$, totaling 14 values.

**Step 3: Run the $y = \pm 2$ timelines.**
If $y = 2$ or $y = -2$:
$$x^2 + (\pm 2)^2 \leq 10$$
$$x^2 + 4 \leq 10 \implies x^2 \leq 6$$
The valid integer values for $x$ are $\{-2, -1, 0, 1, 2\}$.
That's 5 values for $+2$ and 5 values for $-2$, totaling 10 values.

**Step 4: Run the $y = \pm 3$ timelines.**
If $y = 3$ or $y = -3$:
$$x^2 + (\pm 3)^2 \leq 10$$
$$x^2 + 9 \leq 10 \implies x^2 \leq 1$$
The valid integer values for $x$ are $\{-1, 0, 1\}$.
That's 3 values for $+3$ and 3 values for $-3$, totaling 6 values.

**Step 5: Check the outer boundaries ($|y| \geq 4$).**
If $y = \pm 4$, $y^2 = 16$.
$$x^2 + 16 \leq 10 \implies x^2 \leq -6$$
Since a squared real number cannot be negative, there are zero solutions beyond this boundary. The map ends here.

Step 6: Tally the total valid coordinate pairs.
Sum the values from all our locked timelines to get the final score:
$$Total = 7 + 14 + 10 + 6 = 37$$

Final Answer: 37