CEMC Cayley, 2022

Combinatorics Difficulty 3.6 AMC 10/12 Find the answer Canada

At the beginning of the first day, a box contains 1 black ball, 1
gold ball, and no other balls. At the end of each day, for each gold
ball in the box, 2 black balls and 1 gold ball are added to the box;
this means that at the end of the first day, there are 5 balls in the
box. If no balls are removed from the box, how many balls are in the box
at the end of the seventh day?

Pick one

Solution

At the beginning of the first day, the box contains 1 black ball
and 1 gold ball.

At the end of the first day, 2 black balls and 1 gold ball are added, so
the box contains 3 black balls and 2 gold balls.

At the end of the second day, $2 \times 2 = 4$ black balls and $2 \times 1 = 2$ gold balls are added, so the box contains 7 black balls and 4
gold balls.

Continuing in this way, we find the following numbers of balls:

End of Day #	Black Balls	Gold Balls
2	7	4
3	$7 + 4 \times 2 = 15$	$4 + 4 = 8$
4	$15 + 8 \times 2 = 31$	$8 + 8 = 16$
5	$31 + 16 \times 2 = 63$	$16 + 16 = 32$
6	$63 + 32 \times 2 = 127$	$32 + 32 = 64$
7	$127 + 64 \times 2 = 255$	$64 + 64 = 128$

At the end of the 7th day, there are thus $255 + 128 = 383$ balls in the box.

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