1. Calculate the value of $y$

$$
\(y=5 \times 6 \times 14 \times 15\)
$$


We can group these numbers to make it easier:

$$
\(\begin{gathered}
y=(5 \times 6) \times(14 \times 15) \\
y=30 \times 210 \\
y=6,300
\end{gathered}\)
$$

2. Evaluate Quantity A

Find the remainder when 6,300 is divided by 18 :
- Divide 6,300 by 18 :

$$
\(6,300 \div 18=350\)
$$

- Since 6,300 is exactly divisible by $\(18(18 \times 350=6,300)\)$, the remainder is 0 .

3. Evaluate Quantity B

Find the remainder when 6,300 is divided by 40 :
- Divide 6,300 by 40 :

$$
\(6,300 \div 40=157.5\)
$$

- To find the remainder, multiply the whole number part (157) by 40 :

$$
\(157 \times 40=6,280\)
$$

- Subtract from the original total:

$$
\(6,300-6,280=20\)
$$


The remainder is 20 .

Alternative Logic: Factoring
- For 18: $\(18=6 \times 3\)$. In our equation $\(y=5 \times 6 \times 14 \times 15\)$, we have a 6 and a 15 (which is $\(3 \times 5\)$ ). Since $y$ contains both a 6 and a 3 as factors, it must be divisible by 18 with 0 remainder.
- For 40: $\(40=5 \times 8\)$. In our equation, we have a 5 , but do we have an 8 ? The even factors are $\(6(2 \times 3)\)$ and $\(14(2 \times 7)\)$. Together, they provide $\(2 \times 2=4\)$. We are missing one more factor of 2 to make an 8 . Therefore, it cannot be perfectly divisible by 40 , resulting in a non-zero remainder.

Comparison:
- Quantity A: 0
- Quantity B: 20