Concerning cheque: What was the exact amount of that cheque?

Question

A man went into a bank to cash a cheque. In handing over the money the cashier, by mistake, gave him dollars for cents and cents for dollars. He pocketed the money without examining it, and spent a nickel on his way home.

He then found that he possessed exactly twice the amount of the cheque. He had no money in his pocket before going to the bank. What was the exact amount of that cheque?

Answer 1

Assuming the check originally had x dollars and y cents, it was worth 100*x + y cents. The amount he got 100*y + x cents.
After spending 5 cents, he had 100*y + x - 5 cents, which is twice 100*x + y.
So 2(100x + y) = 100y + x - 5 or 200x + 2y = 100y + x - 5 or 199x - 98y = -5 which doesn't look like enough information to solve the problem except that x and y must be whole numbers. so 98y = 199x + 5, y = (199x + 5)/98 = 2x + (3x + 5)/98
Since x and y are whole numbers, so must be (3x + 5)/98.
Call it z = (3x+5)/98 so 98z = 3x + 5, or 3x = 98z - 5 or x = (98z - 5)/3 or x = 32z-1 + (2z-2)/3.
Since everything is a whole number, so must be (2z-2)/3.
Call it w = (2z-2)/3, so 3w = 2z-2 so z = (3w+2)/2 or z = w + 1 + w/2. So w/2 must be whole, or w must be even.
So try w = 2. Then z = 4. Then x = 129. Then y = 262.
These answers are no good, but looking at the original equation, if you decrease y by 199 and x by 98, the answer will still obviously work, so do that: y = 63 and x = 31.
Original check: $31.63 Recieved: $63.31.
Subtract a nickel: $63.26 = 2 times $31.63.

Answer 2

Ciao, here simple code I used on Matlab for finding the answer which it is 31.63 $.

clc
%initialization
ttl0=0;
ttl2=0;
ttl1=0;
add=0.01;
diff=1;

d0=1
c0=0.01
ttl0=d0+c0
%end init

while abs(diff)>0.01
ttl0=ttl0+add;
d0=floor(ttl0);
c0=ttl0-floor(ttl0);

d1=c0*100;
c1=d0/100;
ttl1=d1+c1;
ttl2=ttl1-0.05;

diff=ttl2-(2*ttl0);
end

clc

check=ttl0
double_check=ttl0*2
current_wallet=ttl2

Bye
Marco