Check Cashing Problem

A man writes a check for d dollars and c cents. The clerk, in error, gave him c dollars and d cents. After spending 5 cents, the man finds the remaining money is twice the amount of the check. Determine the amounts.

We could represent the amount of the check as 100d + c and the amount the clerk gave him as 100c + d and determine an equation to be

2(100d + c) = (100c + d) - 5

Simplify:                                         200d +  2c = 100c + d - 5

199d +  5 = 98c

Now, c and d must be integers with both c and d less than 100.  So the question becomes what is the smallest integer  d that will result in an integer value for  c ?      We could klutz around with this equation but it might be a little easier to see if we consider that

199d = 196d + 3d =  2(98)d + 3d

So

Now we can see that 3d + 5 must be a multiple of 98.    If 3d + 5 = (1)(98) we have

3d + 5 = 98    which means that  d = 31   and therefore   c = 62 + 1 = 63.

The check was \$31.63.

Assessing that result, the clerk would have given out \$63.31.   After .05 was spent, the amount of money remaining was \$63.26 which is   2 x \$31.63.    Confirmed -- almost.   We still have to confirm that other values of d would not work.

Check

3d + 5 = (2)(98)
3d + 5 = (3)(98)
3d + 5 = (4)(98)

The first two do not lead to an integer value of  d;   the third one gives a value of   d > 100.

If   3d + 5  is a larger multiple of 98, then d would be more that 100 and so  d = 31, c = 63 is the only solution.

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