sud21 wrote:

There are a few ways to think about this problem, but I'll be honest, one of the easiest is actually going to be to quickly list some possible values of N and look for a pattern:

Since n has a remainder of 3 when divided by 7, it must be exactly 3 larger than any multiple of 7...0, 7, 14, etc... So, let's list the first few:

3, 10, 17, 24, 31 (and now just keep adding 7 to get the next numbers in the list...

I'm going to catalog this stuff in a list/table format so that I can test each statement easily. The first number is the possible value of N, the second in parenthesis is the remainder when divided by 5

3 (3)
10 (0)
17 (2)
24 (4)
31 (1)
38 (3)
45 (0)
52 (2)
59 (4)
66 (1)
73 (3)

Notice that the remainder when divided by 5 has a repeating pattern, 3, 0, 2, 4, 1...

Now let's test each statement:

Statement (1) n is even...
Look at our list, even Ns could be 10, 24, 38, 52...but these each have different remainders, NOT sufficient!

Statement (2) when n is divided by 35, the remainder is 3...
Let's make a quick list where we have the remainder when divided by 35 in [brackets]

3 (3) [3]
10 (0) [10]
17 (2) [17]
24 (4) [24]
31 (1) [31]
38 (3) [3]
45 (0) [10]
52 (2) [17]
59 (4) [24]
66 (1) [31]
73 (3) [3]

Notice that when division by 35 gave a remainder of 3, division by 5 gave a remainder of 3. So this statement is SUFFICIENT!

The correct answer is B.

Hope this helps!

Whit