The five-digit number 45yz0 is divisible by 40. What is the maximum possible value of (y + z)?

The correct option is 16.

To find the maximum possible value of
$y+z$
for the five-digit number
$45yz0$
being divisible by 40, we can analyze the divisibility rules.

Step 1: Understand divisibility by 40
A number is divisible by 40 if and only if it is divisible by both 10 and 4. Since the last digit of
$45yz0$
is 0, the number is naturally divisible by 10. For the number to be divisible by 40, dividing it by 10 must result in a number that is divisible by 4. Thus, the remaining four-digit integer
$45yz$
must be divisible by 4.

Step 2: Divisibility rule for 4
A number is divisible by 4 if the number formed by its last two digits is divisible by 4. Therefore, the two-digit number
$yz$
must be divisible by 4, where
$y$
and
$z$
are single-digit integers from 0 to 9.

Step 3: Maximize the sum of the digits
We want to find the maximum possible value of
$y+z$.
Let us test the largest possible values for the digit
$y$
and find the corresponding values of
$z$
such that
$yz$
is a multiple of 4:

Case 1: Let
$y=9$.
The two-digit numbers starting with 9 that are divisible by 4 are 92 and 96.
- If
$y=9$
and
$z=2$,
then the sum is
$y+z=9+2=11$.
- If
$y=9$
and
$z=6$,
then the sum is
$y+z=9+6=15$.

Case 2: Let
$y=8$.
The two-digit numbers starting with 8 that are divisible by 4 are 80, 84, and 88.
- If
$y=8$
and
$z=0$,
then
$y+z=8$.
- If
$y=8$
and
$z=4$,
then
$y+z=12$.
- If
$y=8$
and
$z=8$,
then
$y+z=8+8=16$.

Case 3: Let
$y=7$.
The two-digit numbers starting with 7 that are divisible by 4 are 72 and 76.
- If
$y=7$
and
$z=6$,
then
$y+z=13$.

Comparing the values, the maximum possible sum is 16, which is achieved when
$y=8$
and
$z=8$.