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?
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.

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Ok
As always, I see another step as I post...
x(a+b+c+d)=1720
Though that gets me no closer :)