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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.

jcsd
Sep23-04, 10:39 AM
bah! I can do it with only 2 4's:

[cosh(4) + 4] = 31

A_I_
Sep23-04, 10:55 AM
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

A_I_
Sep23-04, 11:13 AM
how can we do it on a calculator.. i mean the cosh.. or does it have any formula?

Tide
Sep23-04, 11:17 AM
\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.

Chrono
Sep23-04, 01:19 PM
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.

recon
Sep23-04, 10:53 PM
Is this a brain teaser or a question you found in a mathematics textbook? :smile:

Tide
Sep23-04, 11:54 PM
4^{4-\sqrt 4} = 31

... base 5! :-)

gazzo
Sep24-04, 12:05 AM
Oooh. Does this count?


{\lfloor}\int_{4}^{4+4} cosh (\theta + \theta + \theta){\rfloor}

gazzo
Sep24-04, 12:23 AM
oh, ignore my last post. bah. bad calculator! bad bad!