Work out. 1 1/5 + 2 1/3 m = 2 x 3^a x 7^b and n = 3 x 72 x 11 Given that m > 126, find the smallest values of a and b which make m a multiple of 126. Explain why 33n is a perfect square.

For the prime factor 2: $2^1$ in $m$ matches $2^1$ in 126.
For the prime factor 3: $3^a$ in $m$ means $a \ge 2$.
For the prime factor 7: $7^b$ in $m$ means $b \ge 1$.

If we take the smallest possible values from Step 2 ($a=2, b=1$), then $m = 2 \times 3^2 \times 7^1 = 2 \times 9 \times 7 = 126$.
This value of $m$ is not strictly greater than 126, so we need to increase $m$. To do this, we must increase either $a$ or $b$ by at least 1.

Let's test the smallest possible value for $a$, which is $a=2$.
If $a=2$, then $m = 2 \times 3^2 \times 7^b = 18 \times 7^b$.
We need $18 \times 7^b > 126$.
Divide by 18: $7^b > \frac{126}{18} = 7$.
So, $7^b > 7^1$, which implies $b > 1$.
Since $b$ must be an integer and $b \ge 1$, the smallest integer value for $b$ that satisfies $b > 1$ is $b=2$.
Therefore, the smallest values are $a=2$ and $b=2$.
Let's check this: $m = 2 \times 3^2 \times 7^2 = 2 \times 9 \times 49 = 18 \times 49 = 882$.
882 is a multiple of 126 ($882 = 7 \times 126$) and $882 > 126$.

The final answer is $\boxed{\mathbf{a=2, b=2}}$.

Given $n = 3 \times 7^2 \times 11$.

A number is a perfect square if all the exponents in its prime factorization are even. In the expression $3^2 \times 7^2 \times 11^2$, all the exponents (2, 2, and 2) are even.
Therefore, $33n$ is a perfect square. It can also be written as $(3 \times 7 \times 11)^2 = (231)^2$.

The final answer is $\boxed{\text{Because the prime factorization of } 33n \text{ is } 3^2 \times 7^2 \times 11^2, \text{ where all prime factors have even exponents.}}$