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

Question

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

1. A. 16 (Correct answer)
2. B. 18
3. C. 7
4. D. 5

Correct Answer

Option A — 16

Detailed Solution & Explanation

The correct answer is 16.

Key Points

• $45yz0$ ends in 0, so it is already divisible by 10; for divisibility by 40 the number $\div10 = \overline{45yz}$ must be divisible by 4, i.e. $\overline{yz}$ divisible by 4.
• To maximise $y+z$, take $\overline{yz}=88$ (divisible by 4): $y+z=8+8=16$.

Additional Information

• Divisibility by a composite number is tested through its coprime factors: 40 = 8 × 5, so a number divisible by 40 must be divisible by both.
• The practical tests: divisible by 4 if the last two digits are, by 8 if the last three are, and by 5 if it ends in 0 or 5.
• To maximise a digit sum under a divisibility constraint, start from the largest digit and step down until the constraint is met.