An absentminded bank teller switches the dollars and cents when he cashed a check for Mr. Spencer, giving him dollars instead of cents, and cents instead of dollars. After buying a five cent newspaper, Mr. Spencer discovered he had left exactly twice as much as his original check. What was the amount of the check?

Let x be the dollars and y be the cents of the amount of the check.

(1) 100y + x - 5 = 2(100x+y)
(2) y = (199x+5)/98

To find a solution, both x and y must be integers, and x must be between 0 and 99, inclusive.

So 199x+5 must be a multiple of 98. Being a rather dull guy, I just started searching. I found that when x=0, the remainder is 5. When x=1, the remainder is 8, and each time you increase x, the remainder increases by 3. That is, the remainder is 3x+5.

(If I weren't so dull, I would have rewritten equation 2 as follows)

(3) y = 2x + (3x+5)/98

So the trick becomes finding an integer solution to

3x+5=98 or
3x+5=2*98 or
3x+5=3*98, etc.

(I wrote "etc." but a little thought, and I can see there aren't any more equations to check, because if 3x+5=4*98 for some integer, x, then 3x+5=1*98 for some other integer, x, which is smaller by 98 than the original x.)

It turns out 3x+5=98 has an integer solution, when x=31, and the other equations don't have integer solutions.

When x=31, y=63, so the original check was for $31.63.

Let's check this result.

He cashes the check, and gets $63.31. Then he buys a paper for $0.05, leaving him with $63.26, which is indeed exactly twice $31.63.