, 2016

The first bucket contains 7 red discs.

Each bucket after the first contains 3 more red discs than the previous bucket.

Thus, the second bucket contains $7+3=10$ red discs, the third bucket contains $10+3=13$ red discs, and the fourth bucket contains $13+3=16$ red discs.
Solution 1

Let $b$ represent the number of buckets after the first.

Since the first bucket contains 17 green discs and each bucket after the first contains 1 more green disc than the previous bucket, then there are $17+b$ green discs inside the bucket $b$ after the first.

Since the first bucket contains 7 red discs and each bucket after the first contains 3 more red discs than the previous bucket, then there are $7+3b$ red discs inside the bucket $b$ after the first.

The number of green discs in a bucket is equal to the number of red discs inside the same bucket when $17+b=7+3b$ or when $2b=10$, and so when $b=5$.

Thus, there are an equal number of red discs and green discs in the bucket 5 after the first bucket, which is the 6$^{th}$ bucket.

Solution 2

Using the fact that the first bucket contains 17 green discs and 7 red discs, and each bucket after the first contains 1 more green disc and 3 more red discs than the previous bucket, then we may summarize the number of green and red discs inside each bucket.

Bucket Number
Number of green discs
Number of red discs

1
17
7

2
18
10

3
19
13

4
20
16

5
21
19

6
22
22

Therefore, the 6$^{th}$ bucket contains an equal number of red discs and green discs.

(Note that since the number of red discs is increasing by 3 each bucket and the number of green discs is increasing by 1 each bucket, then this is the only bucket in which the number of red discs and green discs will be equal.)

Solution 3

In the first bucket, there are 17 green discs and 7 red discs. Each bucket after the first contains 1 more green disc and 3 more red discs than the previous bucket.

Since there are 2 more red discs than green being put in, then the difference between the numbers of green and red discs will decrease by 2 for each bucket after the first.

Since the original difference is $17-7=10$, then it takes $10 \div 2 = 5$ more buckets to arrive at a bucket where the numbers of green and red discs will be equal.

Therefore, the 6$^{th}$ bucket contains an equal number of red discs and green discs.
As in part (b), there are $17+b$ green discs and $7+3b$ red discs inside the bucket $b$ after the first.

The number of red discs in a bucket is equal to twice the number of green discs inside the same bucket when $7+3b=2(17+b)$ or when $7+3b=34+2b$, and so when $b=27$.

In the 27$^{th}$ bucket after the first (the 28$^{th}$ bucket), there are $17+27=44$ green discs and $7+3(27)=88$ red discs. (We note that 88 is indeed twice 44.)

The total number of discs in this bucket is $44+88=132$.