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.

Here are the solutions to the problems:

1. Work out $1\frac{1}{5} + 2\frac{1}{3}$

Step 1: Convert the mixed numbers to improper fractions. $1\frac{1}{5} = \frac{1 \times 5 + 1}{5} = \frac{6}{5}$ $2\frac{1}{3} = \frac{2 \times 3 + 1}{3} = \frac{7}{3}$

Step 2: Find a common denominator for the fractions, which is 15. $\frac{6}{5} = \frac{6 \times 3}{5 \times 3} = \frac{18}{15}$ $\frac{7}{3} = \frac{7 \times 5}{3 \times 5} = \frac{35}{15}$

Step 3: Add the fractions. $\frac{18}{15} + \frac{35}{15} = \frac{18 + 35}{15} = \frac{53}{15}$

Step 4: Convert the improper fraction back to a mixed number. $\frac{53}{15} = 3\frac{8}{15}$ The final answer is \boxed{3\frac{8{15}}}.

2. Given $m = 2 \times 3^a \times 7^b$ and $n = 3 \times 7^2 \times 11$

a) Given that $m > 126$, find the smallest values of $a$ and $b$ which make $m$ a multiple of 126.

Step 1: Find the prime factorization of 126. $126 = 2 \times 63 = 2 \times 9 \times 7 = 2^1 \times 3^2 \times 7^1$

Step 2: For $m = 2 \times 3^a \times 7^b$ to be a multiple of $126 = 2^1 \times 3^2 \times 7^1$, the exponents of its prime factors must be greater than or equal to those of 126. 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 $b \ge 1$.

Step 3: Consider the condition $m > 126$. 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.

Step 4: Find the smallest $a$ and $b$ that satisfy $a \ge 2$, $b \ge 1$, and $m > 126$. 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 $\boxed{a=2, b=2}$.

b) Explain why $33n$ is a perfect square.

Step 1: Substitute the expression for $n$ into $33n$. Given $n = 3 \times 7^2 \times 11$. $33n = 33 \times (3 \times 7^2 \times 11)$

Step 2: Express 33 as a product of its prime factors. $33 = 3 \times 11$

Step 3: Substitute the prime factors of 33 into the expression for $33n$ and combine like terms. $33n = (3 \times 11) \times (3 \times 7^2 \times 11)$ $33n = 3^1 \times 11^1 \times 3^1 \times 7^2 \times 11^1$ $33n = 3^{1+1} \times 7^2 \times 11^{1+1}$ $33n = 3^2 \times 7^2 \times 11^2$

Step 4: Explain why the result is a perfect square. 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 $\boxed{Because the prime factorization of 33n is 3^2 \times 7^2 \times 11^2, where all prime factors have even exponents.}$

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