Which of the following number is divisible by both 3 and 8?

Correct option is C

Given:
Options: 4260, 4016, 4128, 3412
Number must be divisible by both 3 and 8.
Formula Used:
Divisibility by 3: Sum of digits is divisible by 3.
Divisibility by 8: Last three digits form a number divisible by 8.
Solution:
Check option A (4260):
Sum of digits = 4 + 2 + 6 + 0 = 12 (Divisible by 3).
Last three digits = 260. $$\frac{260}{8}$$​ = 32.5 (Not divisible by 8).
Check option B (4016):
Sum of digits = 4 + 0 + 1 + 6 = 11 (Not divisible by 3).
Check option C (4128):
Sum of digits = 4 + 1 + 2 + 8 = 15 (Divisible by 3).
Last three digits = 128.  $$\frac{128}{8}$$​ = 16 (Divisible by 8).
Since 4128 satisfies both conditions, it is the correct number.
Final Answer
So the correct answer is (c)