Classic problem: Four 4's

[mfgoode]

93 seems to have stumped all of you!
Well here it is
93=4sq(4!/4)-(4-4/4)
Mally

[Wacky]

Sorry, Mally.
The rules allow ONLY four 4's.

[Ken_g6]

I can do 93 with only 5 4's:

4*4sq+4/.4^2+4

But I haven't figured out 4 4's, yet.

[Wacky]

248 = 4sqsq -4 -4
252 = 4sqsq -4
260 = 4sqsq +4
264 = 4sqsq + 4 + 4
268 = 4sqsq + 4 + 4 + 4
255 = 4sqsq -4/4
251 = 4sqsq -4/4 -4
257 = 4sqsq + 4/4
253 = 4sqsq + 4/4 -4
228 = 4sqsq -4! -4

230 = (4*4! -4) / .4
239 = (4*4! -.4) / .4
240 = 4*4! /.4
236 = (4*4!/.4) - 4
244 = (4*4!/.4) + 4

284 = 4sqsq + 4! + 4
276 = 4sqsq + 4! - 4

262 = 4sqsq + 4!/4
258 = 4sqsq +4!/4 -4

[cheesehead]

Originally Posted by Ken_g6

171 = ((4-4/4)^sq+4)^sq

Correction: 169 = ((4-4/4)^sq+4)^sq

:-)

[Ken_g6]

Originally Posted by cheesehead

Correction: 169 = ((4-4/4)^sq+4)^sq

:-)

Correct. Sometimes I look at the Excel row number, which is 2 off.

175 = (4!+4)/.4^sq
171 = (4!+4)/.4^sq-4

[rogue]

I have an idea for a means to solve some of the remaining values (focusing more on the odd ones).

4 and 24, each require one 4
3, 5, 7, 11, 15, and 17, each require three 4s

Someone could write a program that could do repeated squarings of these numbers until the difference between two of them is 93. It shouldn't be hard to write, but it is possible that the number of squarings would exceed available memory on ones computer.

[xilman]

Originally Posted by rogue

I have an idea for a means to solve some of the remaining values (focusing more on the odd ones).

4 and 24, each require one 4
3, 5, 7, 11, 15, and 17, each require three 4s

Someone could write a program that could do repeated squarings of these numbers until the difference between two of them is 93. It shouldn't be hard to write, but it is possible that the number of squarings would exceed available memory on ones computer.

While it is true that 5 requires three 4s, I observe that 25 = 4/(.4sq). This frees up another 4 to be used elsewhere.

Paul

[fetofs]

Solutions used with the "256-x" trick:

211=4^sq[sup]sq[/sup]-(4!/4)!/4^sq
217=4^sq[sup]sq[/sup]-(4^sq-.4)/.4

Believe it or not,the "subtract from 256" trick worked out on numbers higher than 256!

295=4^sq[sup]sq[/sup]-(.4-4^sq)/.4

Here "256+x" worked:

185=(4-4/.4^sq)^sq-256

EDIT: grandpascorpion really got it right on this one. Another one I found recently:

206=4^sq[sup]sq[/sup]-(4 + 4)/.4^sq

[Ken_g6]

I can't get beyond 2^31 with Excel, but I tried your ideas, and

I got a few more. I may not have found 93, but I independently came up with 185, 211, and 217, which fetofs just posted.

Here are the ones I found earlier that I hadn't listed here yet:
226 = 4^sq[sup]sq[/sup]-4!-4!/4
227 = 4^sq[sup]sq[/sup]-4/.4^sq-4
233 = 4^sq[sup]sq[/sup]+4!-4/4
234 = 4^sq[sup]sq[/sup]-4^sq-4!/4
235 = 4^sq[sup]sq[/sup]-4/.4^sq+4
238 = 4^sq[sup]sq[/sup]-4!+4!/4

If you haven't found 272, you haven't been looking very hard!
272 = 4^sq[sup]sq[/sup]+4^sq
274 = 4^sq[sup]sq[/sup]+4!-4!/4
275 = 44/.4^sq
277 = 4^sq[sup]sq[/sup]+4/.4^sq-4
278 = 4^sq[sup]sq[/sup]+4^sq+4!/4
282 = 4^sq[sup]sq[/sup]+4^sq+4/.4
285 = 4^sq[sup]sq[/sup]+4+4/.4^sq
286 = 4^sq[sup]sq[/sup]+4!+4!/4
288 = 4^sq[sup]sq[/sup]+4^sq+4^sq
289 = (4^sq+4/4)^sq
291 = 44/.4^sq+4^sq
292 = 4^sq[sup]sq[/sup]+4!^sq/4^sq
293 = 4^sq[sup]sq[/sup]+(4!^sq+4^sq)/4^sq
296 = 4^sq[sup]sq[/sup]+4!+4^sq
297 = 4^sq[sup]sq[/sup]+4/.4^sq+4^sq
299 = 44/.4^sq+4!
300 = 4^sq[sup]sq[/sup]+44

[Ken_g6]

I just found one with the 576-x trick.
576-x? 576 = 4!^sq

287 = 4!^sq-(4^sq+4/4)^sq

15 left!