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------------------------------------------------ A box contains only green, red, blue, and yellow counters. There is always at least one green counter amongst any 27 counters chosen from the box, at least one red counter amongst any 25 counters chosen, at least one blue counter amongst any 22 counters chosen, and at least one yellow counter amongst any 17 counters chosen. What is the largest number of counters that could be in the box?

Let's denote the number of green, red, blue, and yellow counters in the box as G, R, B, and Y, respectively. We are trying to find the largest possible value for the sum G + R + B + Y.

The given conditions can be expressed as follows:

There is always at least one green counter amongst any 27 counters chosen: G / 27 ≥ 1

There is always at least one red counter amongst any 25 counters chosen: R / 25 ≥ 1

There is always at least one blue counter amongst any 22 counters chosen: B / 22 ≥ 1

There is always at least one yellow counter amongst any 17 counters chosen: Y / 17 ≥ 1

From these inequalities, we can find the minimum values for each of G, R, B, and Y:

G ≥ 27

R ≥ 25

B ≥ 22

Y ≥ 17

Now, we want to maximize the total number of counters, so we want to maximize G + R + B + Y.

The largest number of counters will be achieved when we have the minimum possible values for each of G, R, B, and Y:

G = 27, R = 25, B = 22, Y = 17

So, the largest possible number of counters in the box is:

27 (green) + 25 (red) + 22 (blue) + 17 (yellow) = 91 counters.