1. Use Vieta's formulas

For the quadratic $x^2-70x+\lambda=0,$ with roots $\alpha,\beta\in \mathbf N$, we have $\alpha+\beta=70, \qquad \alpha\beta=\lambda.$

So we need natural numbers $\alpha,\beta$ whose sum is $70$, and $\lambda=\alpha\beta.$

Also given: $\frac{\lambda}{2}\notin \mathbf N, \qquad \frac{\lambda}{3}\notin \mathbf N.$ That means $\lambda$ is divisible by neither $2$ nor $3$.

2. Find the minimum possible value of $\lambda=\alpha\beta$

Since $\alpha,\beta\in \mathbf N$ and $\alpha+\beta=70$, the product is minimized when the two numbers are as far apart as possible.

So we test small values of $\alpha$:

• If $\alpha=1$, then $\beta=69$, so $\lambda=1\cdot 69=69.$ But $\frac{69}{3}=23\in \mathbf N,$ not allowed.

• If $\alpha=2$, then $\beta=68$, so $\lambda=136.$ This is divisible by $2$, not allowed.

• If $\alpha=3$, then $\beta=67$, so $\lambda=201.$ Since $201$ is divisible by $3$, not allowed.

• If $\alpha=4$, then $\beta=66$, so $\lambda=264,$ divisible by $2$, not allowed.

• If $\alpha=5$, then $\beta=65$, so $\lambda=325.$ Now $325$ is divisible by neither $2$ nor $3$, so this is allowed.

Hence the minimum possible value is $\lambda=325,$ with roots $\alpha=5,\quad \beta=65.$

3. Evaluate the required expression

We need $\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}.$

Substitute $\alpha=5,\beta=65,\lambda=325$:

• $\alpha-1=4 \implies \sqrt{\alpha-1}=2$
• $\beta-1=64 \implies \sqrt{\beta-1}=8$
• $\lambda+35=325+35=360$
• $|\alpha-\beta|=|5-65|=60$

Therefore, \frac{(2+8)(360)}{60}=rac{10\cdot 360}{60}=10\cdot 6=60.

4. Final answer

$\boxed{60}$

5. Comparison with stored correct answer

Stored correct answer = $60$.

Our derived answer also equals $60$, so they agree.