SOLUTION: 120 five-cent coins are arranged in a row. Then in sequence, every second coin is replaced with a ten-cent coin, every third coin is replaced with a 20-cent coin, every fourth coin

Let there be 120 coin-positions numbered 1-120.

Let the set of largest factors be F = {1,2,3,4,5}

For any coin-position numbered k, if the largest factor of k in F is 1,2,3,4,5,
respectively, then the kth coin-position in the end will have a monetary value
of 5,10,20,50,100 cents, respectively. 

For each member k of F, we will enumerate the coin-position-numbers which have k
as their largest factor.

The number of coin-positions that have as their largest factor of k which is in
F is essentially calculated by this method:

the number that have that largest factor in F MINUS the number that have a
greater largest factor on F.

We use the method of inclusion and exclusion, often called the "sieve" formula.

This is the method where we start out counting too many, and so we subtract
some. But then we subtract too many, so we add some back. But we add back too
many.  So we subtract some more.  But we subtract too many.  So we add some
more, etc., and we continue this addition/subtraction process until there is no
more to add or subtract.

In this case our formula is this sequence with alternating signs:



We first enumerate the number of coin positions which have 1 as their largest
factor in F.  This will require all 5 terms of the formula. 

First term:
The number of coin-positions which have factor 1.  That's all 120, since all
have factor 1.
Total = 120

Second term:
The number of coin-positions which have factors 1 and 2.  That's 120/2=60
The number of coin-positions which have factors 1 and 3.  That's 120/3=40
The number of coin-positions which have factors 1 and 4.  That's 120/4=30
The number of coin-positions which have factors 1 and 5.  That's 120/5=24
Total = 60+40+30+24=154

Third term:
The number of coin-positions which have factors 1, 2 and 3.  That's 120/6=20
The number of coin-positions which have factors 1, 2 and 4.  That's 120/4=30
The number of coin-positions which have factors 1, 2 and 5.  That's 120/10=12
The number of coin-positions which have factors 1, 3 and 4.  That's 120/12=10
The number of coin-positions which have factors 1, 3 and 5.  That's 120/15=8
The number of coin-positions which have factors 1, 4 and 5.  That's 120/20=6
Total = 20+30+12+10+8+6=86

Fourth term:
The number of coin-positions which have factors 1, 2, 3 and 4.  That's 120/12=10
The number of coin-positions which have factors 1, 2, 3 and 5.  That's 120/30=4
The number of coin-positions which have factors 1, 2, 4 and 5.  That's 120/20=6
The number of coin-positions which have factors 1, 3, 4 and 5.  That's 120/60=2
Total = 10+4+6+2=22

Fifth term:
The number of coin-positions which have factors 1, 2, 3, 4 and 5.  That's
120/60=2
Total = 2

Substituting in the formula: 120-154+86-22+2 = 32 with largest factor in F as
1.  These 32 end up as they started, with a 5 cent coin.  So their monetary
value is 32∙5 cents = £1.60.

--------------------------------------------------------

Next we enumerate the number of coin positions which have 2 as their largest
factor in F.  This will require 4 terms of the formula.

First term:
The number of coin-positions which have factor 2.  That's 120/2=60.
Total = 60

Second term:
The number of coin-positions which have factors 2 and 3.  That's 120/6=20
The number of coin-positions which have factors 2 and 4.  That's 120/4=30
The number of coin-positions which have factors 2 and 5.  That's 120/10=12
Total = 62

Third term:
The number of coin-positions which have factors 2, 3 and 4.  That's 120/12=10
The number of coin-positions which have factors 2, 3 and 5.  That's 120/30=4
The number of coin-positions which have factors 2, 4 and 5.  That's 120/20=6
Total = 20

Fourth term:
The number of coin-positions which have factors 2, 3, 4 and 5.  That's 120/60=2

Substituting in the formula: 60-62+20-2 = 16 with greatest factor 2.
These 16 end up with a 10-cent coin.  So their value is 16∙10 cents = £1.60.

--------------------------------------------------------

Next we enumerate the number of coin positions which have 3 as their largest
factor in F.  This will require 3 terms of the formula.

First term:
The number of coin-positions which have factor 3.  That's 120/3=40.
Total = 40

Second term:
The number of coin-positions which have factors 3 and 4.  That's 120/12=10
The number of coin-positions which have factors 3 and 5.  That's 120/15=8
Total = 10+8=18
 
Third term:
The number of coin-positions which have factors 3, 4 and 5.  That's 120/60=2
Total = 2

Substituting in the formula: 40-18+2 = 24 with largest factor in F as 3.
These 24 end up with a 20-cent coin.  So their value is 24∙20 cents = £4.80.

--------------------------------------------------------

Next we enumerate the number of coin positions which have 4 as their largest
factor in F.  This will require 2 terms of the formula.

First term:
The number of coin-positions which have factor 4.  That's 120/4=30.
Total = 30

Second term:
The number of coin-positions which have factors 4 and 5.  That's 120/20=6
Total = 6
 
Substituting in the formula: 30-6 = 24 with greatest factor in F as 4.
These 24 coin positions end up with a 50-cent coin.  So their monetary value is 24∙50 cents = £12.00.

----------------------------------

So the total monetary value of all the coins in the end is 
£1.60 + £1.60 + £4.80 + £12.00 + £24.00 = £44.00

Edwin