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December 20, 2014

December 20, 2014

Posted by **LARA** on Tuesday, November 19, 2013 at 9:34pm.

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

- MATHS -
**Reiny**, Tuesday, November 19, 2013 at 10:23pmall 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

- MATHS -
**Soraya**, Wednesday, November 20, 2013 at 7:58pmThis problem is like the "Locker Problem" from "Prime Time, Connections 3."

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