# Puzzle

Hi all

Ok, so I got this puzzle right basically by trial and error, but I suspect it could have been solved algebraically.  Could someone show how this could be done?

"I have 4 denominations of coins which add up to £17.20.  Given I have the same number of coins for each denomination, what coins do I have?"

I got as far as:

ax+bx+cx+dx=1720

Any way to solve without trial and error?

### Ok

As always, I see another step as I post...

x(a+b+c+d)=1720

Though that gets me no closer :)

### So ...

scrap that, read the question - doh!

### On a related note

A new trick I learned, sure it's old hat to you all!

37 squared = 1,369

37-25 = 12

50-37 = 13              13x13 = 169

Therefore

1 2

1 6 9

1, 3 6 9

I think that's sort of neat!

### not sure about algebraically

In pennies not pounds:

x must be a whole number, as must the sum a+b+c+d

a+b+c+d must be greater than or equal to 18p and less than or equal to 180p

1720/x must therefore be a whole number between 18 and 180

hence x must be greater than 9 and less than 96

x can only be 10, 20, 40, 43 or 86

therefore a+b+c+d can only be 172, 86, 43, 40 or 20

only one of those is possible using four different numbers from 1, 2, 5, 20, 50 and 100.