Shortcut Trick
For a number to be divisible by 80, it must be divisible by both 8 and 10.
Divisibility by 10 ⇒ Last digit Y = 0.
Divisibility by 8 ⇒ Last three digits 7X0 must be divisible by 8.
For 7X0, if X = 2, then 720 ÷ 8 = 90 (perfectly divisible).
Minimum X + Y = 2 + 0 = 2.
∴ The correct answer is 2.
Alternate Method
Given: The number is 347XY and it is divisible by 80.
Formula: A number is divisible by 80 if it is divisible by 8 and 10 (where 80 = 8 × 10).
Step 1: Determine Y
⇒ For 347XY to be divisible by 10, the unit digit must be 0.
⇒ Y = 0.
Step 2: Determine X
⇒ The number is now 347X0. For divisibility by 8, the last three digits 7X0 must be divisible by 8.
⇒ Possible values for 7X0: 700, 710, 720, 730, 740, 750, 760, 770, 780, 790.
⇒ 720 ÷ 8 = 90 (Divisible, so X = 2)
⇒ 760 ÷ 8 = 95 (Divisible, so X = 6)
Step 3: Minimum Value
⇒ For minimum X + Y, we take X = 2 and Y = 0.
⇒ X + Y = 2 + 0 = 2.
∴ The correct answer is 2.
Additional Information
Divisibility Rule of 8
A number is divisible by 8 if its last three digits are divisible by 8 or if the last three digits are 000.
Divisibility Rule of 10
A number is divisible by 10 if its units digit is 0.
Composite Divisibility
If a number is divisible by two co-prime numbers a and b, then it is also divisible by their product a × b.