Hey ohhhkevo,
I love that first question! Here's a thought process that is extremely helpful for thinking about divisibility.
Every second number is even (2, 4, 6, 8, 10, etc.)
Every third number is divisible by 3 (3, 6, 9, 12, 15, etc.)
If you take a multiple of 3 and add 1 or 2 to it, it's not "every third" anymore - you've broken it off of that every-third-number cycle. Accordingly, if we take 3x (which is a multiple of 3 as long as x is an integer) and add 2 to it, then it can't be divisible by 3 anymore. Try it with any multiple of 3:
3+2 = 5
6+2 = 8
9+2 = 11
12+2 = 14
15+2 = 17
You'll see that the sums each increase by 3...but they're never divisible by 3. And they can't be - when you start with a multiple of 3, you have to add a multiple of 3 in order for the sum to remain divisible by 3.
Maybe think of it this way - picture a machine stamping every three seconds, and a conveyor belt with steel plates going by every three seconds. As long as they're both on that cycle of three, the stamp will hit the plates each time. If we pause the conveyor belt by 2 seconds and then restart, they'll always miss each other - the stamp will hit on beats 3, 6, 9, 12, 15... and the plates will come every 5, 8, 11, 14, 17, 20... Because they're each clicking every third time, but they're off by a beat, they'll never be on the same cycle.
Can I ask from where you got this question? A would be a correct answer...but a number not divisible by 3 cannot be divisible by 6, so actually both A and C would solve this question. Is it possible you copied one of those choices down incorrectly?
For an even trickier application of this same property, you may be interested in this discussion:
https://www.beatthegmat.com/prime-number ... tml#278895
