# Four 4's: Urgent help needed

• Dec 6th 2007, 12:01 PM
cheyanne
Four 4's: Urgent help needed
Hey all,

I could really use some help with this one:

Using four 4's and any combination of math operations, represent as many of the whole numbers 1-100 as you can.

I found operations that result in: 1, 2, 3, 5, 6, 8, 9, 10, 11, 12, 13, 16, 20, 24, 36, 60, 64, 65, 68, and 72.

I am having a tough time finding the others from 1 through 100. Any help will be greatly appreciated.

Thanks,
cheyanne
• Dec 6th 2007, 12:31 PM
wingless
$\displaystyle \frac {4^4}{4+4} = 32$
$\displaystyle 4*4 + \frac {4}{4} = 17$
$\displaystyle 4*4 - \frac {4}{4} = 15$
$\displaystyle 4*(4+4) - 4 = 28$
$\displaystyle 4 + 4 - \frac {4}{4} = 7$
$\displaystyle 4! + 4 + 4 - 4 = 28$
$\displaystyle 4! + 4 + \frac {4}{4} = 29$
$\displaystyle 4! + 4 - \frac {4}{4} = 27$
$\displaystyle {{4+4} \choose {4}} + 4 = 74$
$\displaystyle {{4+4} \choose {4}} - 4 = 66$
• Dec 6th 2007, 12:38 PM
Soroban
Hello, cheyanne!

Quote:

Using four 4's and any combination of math operations,
represent as many of the whole numbers 1-100 as you can.

This is a classic (very old) problem . . .

Here are some tricks you may wish to explore . . .

. . $\displaystyle \frac{4}{.4} \:=\:10$

. . $\displaystyle \frac{4!}{.4} \:=\:60$

One of the hardest is: .$\displaystyle \frac{4!+ 4.4}{.4} \;=\;71$

One of my students came up with: .$\displaystyle \frac{\left(\dfrac{4}{.4}\right)!}{(4+4)!} \;=\;90$

• Dec 7th 2007, 01:40 AM
earboth
Quote:

Originally Posted by cheyanne
Hey all,

I could really use some help with this one:

Using four 4's and any combination of math operations, represent as many of the whole numbers 1-100 as you can.

I found operations that result in: 1, 2, 3, 5, 6, 8, 9, 10, 11, 12, 13, 16, 20, 24, 36, 60, 64, 65, 68, and 72.

Hello,

here are some of the missing numbers:

$\displaystyle 14=\left(\sqrt{4}+\sqrt{4}\right)^{\sqrt{4}}-\sqrt{4}$

$\displaystyle 18= 4 \cdot 4 + \frac4{\sqrt{4}}$

$\displaystyle 19 = 4! - 4 - \frac44$

$\displaystyle 21 = 4! - 4 + \frac44$

$\displaystyle 23 = 4! - \frac{\sqrt{4} + \sqrt{4}}{4}$

$\displaystyle 25 = \left(4 + \frac44 \right)^{\sqrt{4}}$
• Dec 7th 2007, 06:34 AM
Soroban
Hello, everyone!

Did you know there is an Ultimate Solution to the "Four four's"?

Check this out . . .

$\displaystyle n \;=\;-\log_{\left(\frac{4}{\sqrt{4}}\right)} \left[\log_4\left(\sqrt{\sqrt{\sqrt{\cdots\sqrt{4}}}}\ri ght)\right]$
. . . . . . . . . . . . . . . .