4. 88
4. First, note that \( N > 87 \). This is because if there are 75 green, 12 red, 12 white, and 12 blue balls in the bag, we do not necessarily get balls of at least three different colors when at most 87 balls are drawn (for instance, we may only get green and red balls), and this combination of balls satisfies the condition 'if 100 balls are drawn, we can ensure getting balls of all four colors' because, when 100 balls are drawn, only 11 balls are missing and hence no color can be missing from the balls drawn as there are at least 12 balls of each color.
Next, we show that \( N = 88 \) is enough. The given condition ensures at least 12 balls of each color (for otherwise if there are, say, 11 or fewer red balls, it is possible for all of them to be missing when 100 out of the 111 balls are drawn, contradicting the given condition), so when 88 balls are drawn, 23 are left over and it is impossible to have 2 colors missing (which requires at least 24 balls to be left over). It follows that the answer is 88.
4. There are 111 balls in a box, each being red, green, blue or white. It is known that if 100 balls are drawn, we can ensure getting balls of all four colours. Find the smallest integer $N$ such that if $N$ balls are drawn, we can ensure getting balls of at least three different colours.
(1 mark)
There are 111 balls in a box, each being red, green, blue or white. It is known that if 100 balls are drawn, we can ensure getting balls of all four colours. Find the smallest integer $N$ such that if $N$ balls are drawn, we can ensure getting balls of at least three different colours.
(1 mark)