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

Number 2

[hazelchaz]

24 + 5 - 3

Re: Number 2

[rivka.livejournal.com]

Sorry, with #2 you need to treat the numbers as separate integers, so making 24 is not allowed. I should've specified.

[marnanel.livejournal.com]

I couldn't work out the last one, so I googled for it, and eventually found the answer. I wouldn't ever have got it.

[marnanel.livejournal.com]

((4!)-√4)/.4

[rivka.livejournal.com]

Yeah, that was my reaction too when my father gave me the answer. Never in a million years. I posted it because I'm curious to see if anyone gets it.

[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.

[ailbhe]

Assuming I'm right in thinking "numbers" and "digits" are the same here - Rob got the first quickly 44+44/4 and I got the second quickly 25+4-3 but we're struggling on the third.

[rivka.livejournal.com]

Sorry, with #2 you need to treat the numbers as separate integers, so making 24 is not allowed. Rob is correct for #1.

[ailbhe]

Thanks, that makes a huge difference.

[ailbhe]

I didn't come up with my solution before until Rob pointed out that you might have meant to treat them as digits, not integers...

Then we can't seem to do it! I keep wanting to use brackets!

[rivka.livejournal.com]

You mean, just to control order of operations? You can do that.

[wcg.livejournal.com]

I may come back and work on the first two later. For now, I offer this:

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

INT (e^4 + SQRT(4)/4) works out to INT(55.09) which is 55.

[rivka.livejournal.com]

My Dad says:

"Dear Bill,

Two pieces of bad news. e is a number, not an operator. And the correct answer is exactly 55. (Close only counts in horseshoes and hand grenades.)"

[miscrants.livejournal.com]

No, that does work: INT(EXP(4) + SQRT(4)/4)) = 55
The question does say "a very broad set of mathematical operators", and exp() is probably more usual than counting put-a-decimal-point-in-front-of as an operator rather than a different number.

[johnpalmer.livejournal.com]

I think Rivka's Dad wasn't understanding the use of the INT function.

It does work - but it's probably not the intended answer.

I did find a cute way to get to 54.

Take IV IV IV. Turn one upside down and place it underneath another. You now have IV IX. I * IX = 9, V *IX = 45, 9+45 = 54.

I suppose you could drop the I from the third IV - you'd still have IX, just the I would be half-height! - and use it to add one to the result.

[rivka.livejournal.com]

I think Rivka's Dad wasn't understanding the use of the INT function.

Well, he was having it read aloud to him by me, and I don't know what's meant by INT, so it's entirely possible.

[kcobweb.livejournal.com]

answer to #2 (galagan figured this out, not me):
(5 + (3 / 2) ) x 4 = 26

[nolly.livejournal.com]

...you father listens to Car Talk, eh? (at least two of these have been on there recently.)

[rivka.livejournal.com]

Yep. He's mostly blind and spends a lot of his day listening to NPR.

[(Anonymous)]

I have to say, I'm pretty impressed that you got the first one pretty easily - it took me longer than I liked.

The second was trivial - which I also found interesting, that you found it harder.

The third has been interesting... I've gotten a nonexistent answer ("and then plug 4 into the function that, for n an integer, f(n) returns the nth prime" (2, 3, 5, 7 - 7 is the fourth prime. And I can get to 48 pretty easily. But there's no way I can legitimately claim it's "any of a broad set of mathematical operators". Such a function *exists* - primes are well defined, and well ordered. But it's not a common function.)

[rivka.livejournal.com]

The answer to #3 can be seen here (http://rivka.livejournal.com/547639.html?thread=8788279#t8788279). (Highlight the block of white text.)

As far as the relative difficulty of #1 and #2, I think we probably both think that the one we answered easily was easier, but that doesn't mean that one of us is right.