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Eight Eights

Math brain teasers require computations to solve.

 

Puzzle ID:#1146
Fun:*** (2.48)
Difficulty:** (1.86)
Category:Math
Submitted By:Ramlion**

 

 

 



Arrange eight eights so they add up to one thousand.

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Answer

888 + 88 + 8 + 8 + 8 = 1000

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Comments

Sane**
Mar 16, 2005

Nice.
jntrcs*
Apr 21, 2006

i made it harder than it was i was trying to multiply and divide o well
yoyoyo*af*
May 18, 2006

It is possible with all symbols
(8*8-(8+/*(8++8
or
(8*8+8**8-(8+-8

for example
lessthanjake789*us**
Sep 15, 2006

what about 8888/8 - 888/8 = 1111-111 = 1000. bam.
robban310Ase*
Jan 23, 2007

That's nine eights...
javaguru*us*
Jan 20, 2009

I immediately guessed the answer you were expecting would combine multiple 8's into a single number and figured it out. However there are many solutions just using single 8's with operators.

Here are just a few of them:

( 8 x 8 x 8 - 8 ) * ( 8 + 8 ) / 8 - 8 = 1000
8! / (( 8+8 )/8 * ( 8+8 ) + 8 ) - 8 = 1000
( 8^(SQRT( 8+8 )) - 8 x 8 ) / SQRT( 8+8 ) - 8 = 1000
SQRT( 8+8 )^8 / ( 8 x 8 ) - 8 - 8 - 8 = 1000
SQRT( 8^8 )/SQRT( 8+8 ) - SQRT( 8+8 ) x 8 + 8 = 1000
SQRT( 8+8 )^8 / ( 8 x 8 ) - 8 - 8 - 8 = 1000
SQRT( ( 8+8 )^8 ) / ( 8 x 8 ) - 8 - 8 - 8 = 1000
( 8 + 8 ) x 8 x 8 - SQRT( 8+8 ) x 8 + 8 = 1000
8! / ( 8 + 8 + 8 + 8 + 8 ) - SQRT( 8 x 8 ) = 1000
( 8 + 8 ) x ( 8 x 8 - 8/8 ) - SQRT( 8 x 8 ) = 1000
( 8 + 8 ) x ( 8 x 8 - ( 8 + 8 ) / 8 ) + 8 = 1000
( SQRT( 8 )^8 - ( (SQRT( 8 + 8 ) + 8 ) x 8 )) / SQRT( 8+8 ) = 1000

Jimbo*au*
Apr 02, 2009

This is commonly know as the eight eights problem and belongs to a series of problems (the four fours problem, the five fives problem etc) The idea is to make ever integer (whole number) using eight eights and the mathematical operations.
It gets very interesting when you allow square roots, logarithms to any base, indices etc.
What blew me away completely is when I discovered that there is a GENERAL SOLUTION to these problems. There is a formula that allows you to construct any nominated number using eight eights (or four fours , five fives etc). I can't reproduce it off the top of my head but it involves using multiple square roots and then raising the base to an index constructed from a logarithm. I'll research it again. BTW an interesting teaser which has provoked a plethora of solutions. I think you should have allowed for this non uniqueness in the given answer.

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