# MATHS

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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?

• MATHS -

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

• MATHS -

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

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