1
$\begingroup$

The expression twenty minus five has 15 letters (we ignore spaces) and also has the arithmetic value of 15.

You can use standard math constants, operators and functions, represented by english words. Spaces and parenthesis do not count towards the expression length.

Find the longest such expression.

EDIT: As correctly states in https://puzzling.stackexchange.com/a/6304/7403, no such exists, so only bonus now remains.

As a bonus, find the most interesting such expression.

| improve this question | |
$\endgroup$
  • $\begingroup$ how about a variation where the same word cannot be used more than once. $\endgroup$ – John Meacham Dec 24 '14 at 0:08
7
$\begingroup$

There is no determinate result to this question.

plus twenty minus two: equals 18, 18 letters

plus twenty minus two plus twenty minus two: equals 36, 36 letters

plus twenty minus two plus twenty minus two plus twenty minus two: equals 54, 54 letters

and so on...

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Another answer gave infinity, and it was around 2 minutes earlier. That one should be accepted. $\endgroup$ – mdc32 Dec 20 '14 at 0:03
  • 1
    $\begingroup$ There is no obligation to accept the first correct answer. I suggest people accept whatever answer they think is best. $\endgroup$ – Julian Rosen Dec 20 '14 at 2:29
  • 2
    $\begingroup$ I would say, the answer which best explains things should become the accepted answer. $\endgroup$ – BmyGuest Dec 20 '14 at 8:10
  • $\begingroup$ And I think infinity is not a correct answer. The correct answer does not exist. $\endgroup$ – mirelon Dec 20 '14 at 10:13
3
$\begingroup$

4 + (1x21) = 25

Four plus (one times twenty one) = 25 characters

Then continue adding + (1x21) to the equation:

4 + (1x21) + (1x21) = 46

Four plus (one times twenty one) plus (one times twenty one) = 46 characters

4 + (1x21) + (1x21) + (1x21) = 67

Four plus (one times twenty one) plus (one times twenty one) plus (one times twenty one) = 67 characters.

You can keep doing this forever, so the answer is infinity.

| improve this answer | |
$\endgroup$
3
$\begingroup$

The "most interesting" is a little bit subjective. I will try my best, though.

the square root of six to the fourth $=36$

minus e to the i pi plus two to the fourth times four minus two $=63$*

one fourth of six to the third minus eleven $=43$

More will surely follow when found.

*Follow order of operations and this evaluates to 1 + (4*16) - 2

| improve this answer | |
$\endgroup$
2
$\begingroup$

So I'll start with this:

One billion to the one billionth power has 32 characters. We'll call this string POWER. We'll refer to the value 1000000000^1000000000 as the constant BTB.

plus three minus two minus one has 25 characters. This evaluates to 0. We'll call this string MINUS.

plus one plus six plus seven plus fifty has 32 characters. This evaluates to 64. We'll call this string PLUS

If we place the POWER string, contact the MINUS string a number of times equal to BTB / 25 then we will have a string of length BTB + 32 and a value of BTB. We then add the string PLUS to the end, and we have a length of BTB + 64 and a value of BTB + 64.

For the sake of stack exchange's servers, I have only defined the string, not actually written the string out.

You can repeat this process ad infinitum by tacking to the twenty billionth power to the end of the POWER string, and adding the appropriate number of MINUS strings to compensate, since they both have 25 CAREacaters (that's characters we care about. Clever, no?) Remember that now that BTB has increased, we still put the MINUS string BTB times, but we also remove an additional MINUS string for every additional to the twenty billionth power because we don't need to compensate for that anymore.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Added bonus: this is probably extendable pretty easily - one power of 10 string with x number of characters, then another string with value 2x and x characters. $\endgroup$ – mdc32 Dec 19 '14 at 23:56
  • $\begingroup$ Yep, I was busy working on that. I chose one that conveniently fit into 25 characters so it could fit into the framework of using my MINUS string. $\endgroup$ – corsiKa Dec 20 '14 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.