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Home » Ultimates Hack: The Hidden Feature Everyone’s Missing (And You Will Too!) - Treasure Valley Movers

Ultimates Hack: The Hidden Feature Everyone’s Missing (And You Will Too!) - Treasure Valley Movers

📅 April 17, 2026 👤 scraface

Apr 17, 2026

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📌 r = \frac{84}{21} = 4

📌 $$Question: What is the largest possible value of $\gcd(a,b)$, if the sum of two positive integers $a$ and $b$ is $1000$ and both $a$ and $b$ are divisible by a perfect square greater than $1$?

📌 Let $d = \gcd(a,b)$. Then we can write $a = dx$, $b = dy$, where $\gcd(x,y) = 1$. Since $a + b = 1000$, we have:

📌 dx + dy = d(x + y) = 1000 \Rightarrow d \mid 1000

📌 So $d$ must be a divisor of 1000. Also, both $a$ and $b$ must be divisible by a perfect square greater than 1. Since $a = dx$ and $b = dy$, both $dx$ and $dy$ must be divisible by a square $s > 1$. That means $d$ must contain all the square factors of $s$, or at least $x$ and $y$ must together contain the square factors.

📌 To maximize $d$, we look at the largest divisor of 1000 such that $x = \frac{a}{d}$ and $y = \frac{b}{d}$ are coprime and at least one of $x$ or $y$ is divisible by a square greater than 1.

📌 The prime factorization of 1000 is:

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3a + b = 5 \quad \text{(Equation 4)}
Next, use Equations 2 and 3:
4a + 2b + c &= 11 \quad \text{(Equation 2)} \\
9a + 3b + c &= 18 \quad \text{(Equation 3)}
Subtract Equation 2 from Equation 3:
5a + b = 7 \quad \text{(Equation 5)}
Now, solve Equations 4 and 5:
3a + b &= 5 \quad \text{(Equation 4)} \\
Subtract Equation 4 from Equation 5:
2a = 2 \\

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