Math puzzles.

[rivka]

My father brought three math puzzles down with him this weekend. I got the first one pretty easily, struggled with the second, and had no chance in hell in getting the third. So I pass them all along you to guys.

1. Write the number 4 five times, in combination with any of the basic operators, to produce the sum of 55.

2. Take the numbers 2, 3, 4, and 5. Take any of the four basic operators - but you may only use each one once. Produce the sum of 26. Edited to add: You may not put two numbers next to each other to make a larger number (e.g., 25 + 4 -3). Treat them as separate integers.

3. Write the number 4 three times, employing any of a very broad set of mathematical operators, to produce the sum of 55.

Assume that there will be spoilers in the comments section.

Comments

[matthewwdaly.livejournal.com]

Unfortunately, I heard the problem earlier this week (presumably from the same place as your dad), so I can't say for sure. But, while feverishly trying to find a second solution by paper I briefly had an Aha! moment and realized that I had recreated it. It's very aggravating because you can get 48, 54, 60, and 64 all with two 4's, but there's no way to close the gap with the last one without massive cheating.

But, with massive cheating, 55 = (4!)/.4 - sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(4))))))))))))))))))

Better yet, but still not quite

[matthewwdaly.livejournal.com]

It hit me while drifting off to sleep.

55 = C(-4/.4, sqrt(4))

It's a mild abuse of the combinatorial operator to admit negative numbers, but I think most anyone would admit that this should equal (-10)(-11)/2 = 55. Of course, more honestly, 55 = C(11,2) = C(12,3)/4, but you just miss being able to get those in three 4's. A most vexing problem.

Re: Better yet, but still not quite

[johnpalmer.livejournal.com]

All combinations of 0 and less than 0 are defined as 0. That doesn't work. It *is* a good bit of thinking, though.