A small coin nano-challenge

[brickviking]

(03-23-2019 04:05 AM)Gerson W. Barbosa Wrote:  

(03-23-2019 02:21 AM)Paul Dale Wrote:  You found a nickel (5c), thought you had a quarter (25c) at home but really had a dime (10c).

Any currency with three coins A, B and C such that (A + B) = 2(A + C) would work. I.e. A = B - 2C.

Coincidentally, the same choice of letters here, except mine were lower-case.

a: found coin
b: coin imagined to have at home
c: actual coin at home

{ a + b = x; a + c = x/2 }

{ a + b = x; 2a + 2c = x }

->

a + b = 2a + 2c

a = b - 2c

Considering current New Zealand set of coins is { 0.10, 0.20, 0,50, 1.00, 2.00 }, then { a = 1.00, b = 2.00, c = 0.50 } is the only solution for that set of coins.

{a = 0,05; b = 0,25; c = 0,10 } is a solution here (in Brazil) as well.

Just to set the record straight, I am from New Zealand, and there is (as I found) two sets of coins (and not just the one I thought of) that match that. I found a ten cent piece. I thought I had a fifty cent piece at home, but I only had twenty cents. That (of course) made thirty cents, not sixty cents. Sigh.

Inaccurate guess: a=0.50, b=0.10, c=0.60
Actual result: a=0.20, b=0.10, c=0.30 (a is coin at home, b is coin I found, c is total).

Still, thanks for the illumination, everyone. Does anyone know of a currency where there's only one solution?

(Post 329)