How much is the integer part of the following number:

$$
\sqrt{6+\sqrt{6+\ldots+\sqrt{6}}}+\sqrt[3]{6+\sqrt[3]{6+\cdots+\sqrt[3]{6}}}
$$

where both the number of square root and cube root symbols is 100?
Let the first term of the examined sum be denoted by $A$, and the second term by $B$. Clearly,

$$
\sqrt{6}4$, hence the integer part of $A+B$ is 4.

Remarks. 1. It can be seen from the proof that the number of radicals in $A$ and $B$ is irrelevant to the result.

2. The inequality $\sqrt{6}+\sqrt[3]{6}>4$ can be proven, for example, by rearranging it to $\sqrt[3]{6}>4-\sqrt{6}$ and then multiplying both sides by $4+\sqrt{6}$:

$$
\sqrt[3]{6}(4+\sqrt{6})>4^{2}-6=10
$$

this inequality is true because the first factor on the left side is greater than $\frac{10}{6}$, and the second factor is greater than 6.