Daniel begins with 64 coins in his coin jar. Each time he reaches into the jar, he removes half of the coins that are in the jar. How many times must he reach in and remove coins from his jar so that exactly 1 coin remains in the jar?
(A) 5
(B) 32
(C) 6
(D) 7
(E) 63
Each time Daniel reaches into the jar, he removes half of the coins that are in the jar.

Since he removes half of the coins, then the other half of the coins remain in the jar.

We summarize Daniel's progress in the table below.

| Number of Times Coins are Removed | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Number of Coins Remaining in the Jar | 64 | 32 | 16 | 8 | 4 | 2 | 1 |

For exactly 1 coin to remain in the jar, Daniel must reach in and remove coins from the jar 6 times.

ANSWER: (C)