24 1c coins are set out in a row on a table
Then every 2nd coin is replaced by a 2c coin
every 3rd coin is replaced by a 5c coin
every 4th coin is replaced by a 10c coin
every 5th coin is replaced by a 20c coin
every 6th coin is replaced by a 50c coin and every 7th coin is replaced by a $1coin
After all the exchanges have been carried out, what is the total amount left on the table?
The answer is $6.30
all multiples of 7 are $1
7, 14, 21 ------- $3.00
all multiples of 6 are .50
6,12,18,24,------- 4(.50) = $2.00
all multiples of 5 are .20
5,10,15,20 ---- 4(.20) = $.80
all multiples of 4 are .10
4,8,12,16,20,24
but the 12, 20 and 24 have been changed, so
only the 4,8, and 16 contain .10 ---- $.30
only multiples of 3 that have not been changed are
3 and 9 at .05 ------- $ .10
only the 2 and 11 of all the evens is still at .02
----------- $.04
leaving the remaining
1, 11, 13, 17, 19, and 23 to still have their .01
-------- .06
3 + 2 + .8 + .3 + .1 + .04 + .06
= 6.30
you are correct
This problem is like the "Locker Problem" from "Prime Time, Connections 3."
To solve this problem, we need to go through each step of the exchange and calculate the value of the coins on the table after each exchange. Let's break it down:
1. Initially, we have 24 1c coins on the table. The total value of these coins is 24 x 1c = 24c.
2. Every 2nd coin is replaced by a 2c coin. This means we replace coins at positions 2, 4, 6, 8, ..., 24. After this exchange, we have 12 1c coins and 12 2c coins. The total value of the coins at this stage is 12 x 1c + 12 x 2c = 12c + 24c = 36c.
3. Every 3rd coin is replaced by a 5c coin. We replace coins at positions 3, 6, 9, ..., 24. After this exchange, we have 4 1c coins, 12 2c coins, and 8 5c coins. The total value of the coins at this stage is 4 x 1c + 12 x 2c + 8 x 5c = 4c + 24c + 40c = 68c.
4. Every 4th coin is replaced by a 10c coin. We replace coins at positions 4, 8, 12, ..., 24. After this exchange, we have 4 1c coins, 9 2c coins, 8 5c coins, and 5 10c coins. The total value of the coins at this stage is 4 x 1c + 9 x 2c + 8 x 5c + 5 x 10c = 4c + 18c + 40c + 50c = 112c.
5. Every 5th coin is replaced by a 20c coin. We replace coins at positions 5, 10, 15, 20. After this exchange, we have 4 1c coins, 9 2c coins, 8 5c coins, 4 10c coins, and 4 20c coins. The total value of the coins at this stage is 4 x 1c + 9 x 2c + 8 x 5c + 4 x 10c + 4 x 20c = 4c + 18c + 40c + 40c + 80c = 182c.
6. Every 6th coin is replaced by a 50c coin. We replace coins at positions 6, 12, 18. After this exchange, we have 4 1c coins, 9 2c coins, 8 5c coins, 4 10c coins, 3 20c coins, and 3 50c coins. The total value of the coins at this stage is 4c + 18c + 40c + 40c + 60c + 150c = 312c.
7. Every 7th coin is replaced by a $1 coin. We replace coins at positions 7, 14, 21. After this exchange, we have 4 1c coins, 9 2c coins, 8 5c coins, 4 10c coins, 3 20c coins, 2 50c coins, and 2 $1 coins. The total value of the coins at this stage is 4c + 18c + 40c + 40c + 60c + 200c + 200c = 562c.
Converting the total value to dollars, we have 562c = $5.62.
Therefore, the total amount left on the table after all the exchanges is $5.62.