Check Cashing Problem

A man writes a check for

ddollars andccents. The clerk, in error, gave himcdollars anddcents. After spending5cents, 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 + cand the amount the clerk gave him as100c + dand determine an equation to be

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

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 that199d = 196d + 3d = 2(98)d + 3d

So

Now we can see that

3d + 5must be a multiple of98. If3d+ 5 = (1)(98) we have

3d + 5 = 98which means thatd = 31and thereforec = 62 + 1 = 63.The check was

$31.63.Assessing that result, the clerk would have given out

$63.31. After.05was spent, the amount of money remaining was$63.26which is2 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 = 63is the only solution.

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