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
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
2 answers
This problem is like the "Locker Problem" from "Prime Time, Connections 3."
2 answers