To solve for the value of the check you must take into account that the teller transposed the dollars and cents. (e.g. gave you 41.56 when the check was supposed to be 56.41). This means that you must represent the dollars and cents as different variables within the same number (e.g. $d.c) How to do this? Well, $1 = 100 cents, so set everything equal to cents.

Therefore the correct check is (x*100) + y cents, where x is the dollar amount and y the cents.

The cashier gave you (y*100) + x cents, because the numbers were transposed.

With the information in the problem, you know that you spent 1.41 or 141 cents of that incorrect amount, and you are left with 5 times that correct amount. Plug everything into an equation:

**100y + x - 141 = 5 (100x + y)** * incorrect amount - spent = 5 times correct amount*

**100y + x -141 + 141 = 500x + 5y + 141**

**100y + x = 500x + 5y + 141**

**95y = 499x + 141**

**y = (499x + 141)/95**

There would be an infinite number of possibilities, except that you know some information about x and y. Because each at one point represents the cents part of a monetary figure, they can only be between 00 and 99. Also, y, the dollar amount of the payment received,is at least 5 times bigger than x. Also, y is a whole number so, looking at the equation, 499x + 141 has to be divisible by 95 evenly.

Just looking at units digits, only x values ending in 1 or 6 are eligible; otherwise 499x+141/95 doesn't yield an integer. And only 1, 6, 11, 16 qualify, because y is five times bigger than x but less than 99. So plug in each:

**499(1) + 141/95 = 2.344355**

**499(6) + 141/95 = 33. You can plug in to the original equations to get x, and put x and y together to create the value of the original check. $6.33)**