A seven-digit number 489y5z6 is divisible by 72. Which of the options gives the highest possible product of y and z?

The correct option is 42.

We are given a seven-digit number $N=489y5z6$ which is divisible by 72.

Since $72=8\times 9$ (where 8 and 9 are co-prime), the number $N$ must be divisible by both 8 and 9.

Step 1: Test divisibility by 8
A number is divisible by 8 if the number formed by its last three digits is divisible by 8.
The last three digits of $N$ are $5z6$ (which represents $500+10z+6$).

Let's check the possible values of $z$ (where $z$ is a single digit, $0\le z\le 9$):
If $z=1$, 516 ÷ 8 = 64.5 (Not divisible)
If $z=3$, 536 ÷ 8 = 67 (Divisible, so $z=3$ is valid)
If $z=7$, 576 ÷ 8 = 72 (Divisible, so $z=7$ is valid)

Thus, the possible values for $z$ are 3 and 7.

Step 2: Test divisibility by 9
A number is divisible by 9 if the sum of its digits is divisible by 9.
Sum of the digits of $N$:
$4+8+9+y+5+z+6=32+y+z$

Now, let's find $y$ for each possible value of $z$:

Case 1: When $z=3$
Sum of digits = $32+y+3=35+y$
For $35+y$ to be divisible by 9, $y$ must be 1 (since 36 is divisible by 9).
Product of $y$ and $z$ = $1\times 3=3$.

Case 2: When $z=7$
Sum of digits = $32+y+7=39+y$
For $39+y$ to be divisible by 9, $y$ must be 6 (since 45 is divisible by 9).
Product of $y$ and $z$ = $6\times 7=42$.

Comparing the products, the highest possible product of $y$ and $z$ is 42.