View Full Version : 31
turkldo
Sep23-04, 10:24 AM
How can I use 4, 4, and 4 to result in the number 31? You can perform any operation the 4's, but you can only use three 4's.
bah! I can do it with only 2 4's:
[cosh(4) + 4] = 31
what do u mean by cosh.. i didn't understand the h?
turkldo
Sep23-04, 10:56 AM
You are required to use all three of the 4's. I should have been more specific...sorry.
turkldo
Sep23-04, 11:04 AM
AI,
JCSD used the Hyperbolic Cosine function.
The cosh function operates element-wise on arrays. The function's domains and ranges include complex values. All angles are in radians.
Y = cosh(X) returns the hyperbolic cosine for each element of X.
Now, cosh 4 = 27.308
how can we do it on a calculator.. i mean the cosh.. or does it have any formula?
\cosh x = \frac {e^x + e^{-x}}{2}
turkldo
Sep23-04, 11:22 AM
I have a TI-85. On it you select 2nd, Math. Next you press F4 for the HYP menu. Then you choose the type of hyperbolic function you want to use.
I have a TI-85. On it you select 2nd, Math. Next you press F4 for the HYP menu. Then you choose the type of hyperbolic function you want to use.
Or you could just type in (e^x + e^-x)/2
JasonRox
Sep23-04, 02:11 PM
4^0 + 4^0 ... + 4^0 = 31
Technically I didn't use any four's since it is exponent 0.
It is possible I know that.
You have many options to work with.
Is this a brain teaser or a question you found in a mathematics textbook? :smile:
4^{4-\sqrt 4} = 31
... base 5! :-)
Oooh. Does this count?
{\lfloor}\int_{4}^{4+4} cosh (\theta + \theta + \theta){\rfloor}
oh, ignore my last post. bah. bad calculator! bad bad!
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