If n is a positive integer such that $(\sqrt[7]{10})(\sqrt[7]{10})^{2} \ldots(\sqrt[7]{10})^{n}>999$, then the smallest value of n is

We need to find the smallest positive integer $n$ such that the product $(\sqrt[7]{10})(\sqrt[7]{10})^{2} \ldots(\sqrt[7]{10})^{n} > 999$.

---

The expression $(\sqrt[7]{10})(\sqrt[7]{10})^{2} \ldots(\sqrt[7]{10})^{n}$ can be written using exponent notation as:

$(10)^{1/7} \times (10)^{2/7} \times \dots \times (10)^{n/7}$

Since $\sqrt[7]{10} = 10^{1/7}$, $(\sqrt[7]{10})^2 = 10^{2/7}$, and so on.

---

When multiplying powers with the same base, we add the exponents:

$(10)^{1/7} \times (10)^{2/7} \times \dots \times (10)^{n/7} = (10)^{1/7 + 2/7 + \dots + n/7}$

Now we need to find the sum: $\dfrac{1}{7} + \dfrac{2}{7} + \dots + \dfrac{n}{7}$

Factoring out $\dfrac{1}{7}$:

$\dfrac{1}{7} + \dfrac{2}{7} + \dots + \dfrac{n}{7} = \dfrac{1}{7}(1 + 2 + \dots + n)$

Using the arithmetic series formula: $1 + 2 + 3 + \dots + n = \dfrac{n(n+1)}{2}$

Therefore: $\dfrac{1}{7} \times \dfrac{n(n+1)}{2} = \dfrac{n(n+1)}{14}$

---

Our inequality becomes:

$(10)^{n(n+1)/14} > 999$

Since $999 \approx 1000 = 10^3$, we can write:

$(10)^{n(n+1)/14} > 10^3$

Comparing exponents since bases are equal:

$\dfrac{n(n+1)}{14} > 3$

Therefore: $n(n+1) > 42$

---

We need the smallest positive integer $n$ such that $n(n+1) > 42$.

Testing values:

For $n = 5$: $5 \times 6 = 30 < 42$

For $n = 6$: $6 \times 7 = 42$

Since we need $n(n+1) > 42$, not $n(n+1) \geq 42$, let us be more precise with $n = 6$:

$n(n+1) = 42$, so $\dfrac{n(n+1)}{14} = 3$

This gives $(10)^3 = 1000$

Since $1000 > 999$

---

Answer: 6