Question 1164209
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Let d and c be the numbers of dollars and cents on the original check.<br>
Then the value of the check (in cents) is 100d+c.<br>
The amount of money the bank teller gave her had the dollars and cents reversed; the amount of money Susan got was 100c+d.<br>
After spending $3.50, the amount Susan had left was 100c+d-350.<br>
That amount was twice the amount of the original check:<br>
{{{100c+d-350 = 2(100d+c)}}}<br>
This is a Diophantine equation -- a single equation in two unknowns.  It can be solved for a distinct set of answers (in many cases, such as this, a single answer) because of the fact that the unknowns must have integer (often positive integer) values.<br>
In this example, we know c and d are both positive integers less than 100.<br>
Solve the equation for one variable in terms of the other.<br>
{{{100c+d-350 = 200d+2c}}}
{{{98c = 199d+350}}}
{{{c = (199d+350)/98}}}
{{{c = 199d/98+350/98}}}<br>
Perform the divisions on the right to get whole numbers plus remainder fractions.<br>
{{{c = 196d/98+3d/98+294/98+56/98 = (196d/98+294/98)+(3d/98+54/98)}}}
{{{c = (2d+3)+(3d/98+56/98) = (2d+3)+(3d/98)+(4/7)}}}<br>
Now c is an integer; and d is an integer, so 2d+3 is an integer.  That means the expression<br>
{{{3d/98+4/7}}}<br>
must be an integer.<br>
The "first" time it will be an integer is when 3d/98 is equal to 3/7:<br>
{{{3d/98 = 3/7}}}
{{{3d/98 = 42/98}}}
{{{3d = 42}}}
{{{d = 14}}}<br>
Then with d=14,<br>
{{{c = (2d+3)+(3d+56)/98 = (2(14)+3)+98/98 = 31+1 = 32}}}<br>
ANSWER: the amount of the check was $14.32<br>
CHECK:
Original amount: $14.32
Amount teller gave: $32.14
Amount spent: $3.50
Amount left: $28.64
Twice the amount of the check: $28.64<br>
The Diophantine equation has other solutions; however, none of the others satisfy the requirement that both c and d are positive integers less than 100.<br>