Let α, β ∈ N be roots of equation x² - 70x + λ = 0, where λ/2, λ/3 ∉ N. If λ assumes the minimum pos

Q: Wait, why do we reject $\alpha = 1$?

Great question! If $\alpha = 1$, then $\beta = 70 - 1 = 69$. Now, check the divisibility: $69 = 3 \times 23$. Since $69$ is a multiple of $3$, the product $\lambda = \alpha \beta = 69$ would be divisible by $3$. The problem explicitly states that $\frac{\lambda}{3} \notin \mathbb{N}$, so $\alpha = 1$ is disqualified.

Q: Why does minimizing $\alpha$ minimize $\lambda$?

Think of the product $\lambda = \alpha \beta$ with $\alpha + \beta = 70$ as $\lambda = \alpha(70 - \alpha) = 70\alpha - \alpha^2$. This is a downward-opening parabola. The minimum value of this function on the interval $[1, 69]$ occurs at the boundaries, i.e., when $\alpha$ is as small as possible or as large as possible. Since we want the minimum $\lambda$, we look for the smallest valid $\alpha$.

Q: Is there a faster way than testing values?

You could use the discriminant $D = b^2 - 4ac = 70^2 - 4\lambda$. For $\alpha, \beta$ to be integers, $D$ must be a perfect square. However, given the specific divisibility constraints, testing values is actually the most efficient and least error-prone method. It forces you to engage with the number theory behind the problem, which is the heart of the challenge!

Question

Let $\alpha, \beta \in N$ be roots of equation $x^2 - 70x + \lambda = 0$, where $\frac{\lambda}{2}, \frac{\lambda}{3} 
otin N$. If $\lambda$ assumes the minimum possible value, then $\frac{(\sqrt{\alpha-1} + \sqrt{\beta-1})(\lambda + 35)}{|\alpha - \beta|}$ is equal to:

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are presented with the quadratic equation $x^2 - 70x + \lambda = 0$. Let the roots of this equation be $\alpha$ and $\beta$, where $\alpha, \beta \in \mathbb{N}$.

From Vieta's formulas, we establish the following relationships:
$$\alpha + \beta = 70$$
$$\alpha \beta = \lambda$$

The problem imposes a strict constraint: $\frac{\lambda}{2}, \frac{\lambda}{3} 
otin \mathbb{N}$. This implies that $\lambda$ is not divisible by $2$ and not divisible by $3$.

### The Sieve of Divisibility

Since $\lambda = \alpha \beta$, the condition that $\lambda$ is not divisible by $2$ or $3$ implies that neither $\alpha$ nor $\beta$ can be divisible by $2$ or $3$. If either root were a multiple of $2$ or $3$, their product $\lambda$ would necessarily inherit that divisibility, violating the given constraint.

We must find a pair $(\alpha, \beta)$ such that $\alpha + \beta = 70$, where both $\alpha$ and $\beta$ are coprime to $2$ and $3$.

### The Search for the Minimum

To minimize $\lambda = \alpha \beta$ for a fixed sum, we must select $\alpha$ and $\beta$ to be as far apart as possible. We test values of $\alpha$ starting from the smallest natural numbers:

1. If $\alpha = 1$, then $\beta = 69$. Since $69 = 3 	imes 23$, it is divisible by $3$. **Rejected.**
2. If $\alpha = 2, 3, 4$, these are either divisible by $2$ or $3$. **Rejected.**
3. If $\alpha = 5$, then $\beta = 70 - 5 = 65$.

Checking the divisibility for $\alpha = 5$ and $\beta = 65$:
- $5$ is not divisible by $2$ or $3$.
- $65$ is not even (not divisible by $2$) and the sum of its digits is $11$ (not divisible by $3$).

Both roots satisfy the condition. Thus, the minimum value is:
$$\lambda = 5 	imes 65 = 325$$

### The Final Evaluation

We now substitute $\alpha = 5$, $\beta = 65$, and $\lambda = 325$ into the target expression:
$$\frac{(\sqrt{\alpha-1} + \sqrt{\beta-1})(\lambda + 35)}{|\alpha - \beta|}$$

First, calculate the numerator components:
$$\sqrt{5-1} + \sqrt{65-1} = \sqrt{4} + \sqrt{64} = 2 + 8 = 10$$
$$\lambda + 35 = 325 + 35 = 360$$

Next, calculate the denominator:
$$|\alpha - \beta| = |5 - 65| = |-60| = 60$$

Combining these results:
$$\frac{10 	imes 360}{60} = \frac{3600}{60} = 60$$

The final resulting answer is **60**.

Animated Solution for Mathematics - Quadratic Equations: Let $α, β \in N$ be roots of equation $x^{2} - 70 x + λ = 0$ , where $\frac{λ}{2}, \frac{λ}{3} \in / N$ . If $λ$ assumes the minimum possible value, then $\frac{( α - 1 + β - 1 ) ( λ + 35 )}{∣ α - β ∣}$ is equal to:

Enter Numerical Value:

Visualized Solution

Identify Root Properties

Analyze Divisibility Constraints

Strategy to Minimize $λ$

Test $α = 1$

Skip Multiples of $2$ and $3$

Validate $α = 5$

Calculate Minimum $λ$

Set up Final Expression

Evaluate Square Roots

Calculate Numerator Factor

Evaluate Denominator and Final Computation

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Sieve of Divisibility

The Search for the Minimum

The Final Evaluation

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Enter Numerical Value:

Visualized Solution

Identify Root Properties

Analyze Divisibility Constraints

Strategy to Minimize .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Test .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=1

Skip Multiples of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3

Validate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=5

Calculate Minimum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ