Verified answer

The first number divisible by 6 is 36 and the last is 390

This an arithmetic sequence with a{1}=36, d=6

a{n} = a{1} + (n-1)d

390 = 36 + (n-1)6

Therefore n=60

We can check to find that neither 32 nor 395 are divisible by 6.

The first 6-multiple after 32 is 36 ( = 6 x 6).

The next one is 42 ( = 6 x 7), then 48 ( = 6 x 8), and so on

This continues up to the last 6-multiple before 395, which is 390 ( = 6 x 65)

So, all the integers in the range can be listed as 6 times these multiples:

6, 7, 8, ..., 63, 64, 65

If we count these starting from 1, notice it's the same as subtracting 5 from each number:

1, 2, 3, ..., 58, 59, 60

And now it's easy to see that there are 60 of them.

Use the divisibiilty test

6 if the number is even and divisible by 3.

Example the number is 534:

Rule

3

= 5 + 3 + 4 = 12, and 12 is divisible by 3

6

= 534 is EVEN and divisible by 3

To get you started (Look at the numbers that are even only, the rest isn't important)

36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, you do the rest

Notice all the numbers are even and if added together is divisible by 3