# Short maths problem

• August 11th 2009, 08:54 PM
bruxism
Short maths problem
Hi everyone.

I'm currently tutoring a 15 year old maths student and his teacher has set an assignment question that's a bit wierd.

'Using four 6's and the operations (plus, minus, divide, multiply and brackets) make equations equal to the numbers 1 through 9)

The following examples are given

66/66 = 1
6/6 + 6/6 = 2
(6 + 6 + 6)/6 = 3

I've got answers for all except for 9. Anyone know how you could make an equation equal to nine?

Secondly, this is supposed to be a piece of assessment testing basic algebra. I really don't see how this problem can be done algebraically. Am i missing the point of what this teacher is trying to do here?
• August 11th 2009, 10:22 PM
garymarkhov
Yeah, 0 through 8 are straightforward, even 10 and 12 aren't a problem. I can't figure out 9! I am hesitant to say it's impossible, because plenty of puzzles that look that way turn out not to be... but I wonder. I sure would like to see if someone comes up with the answer!
• August 12th 2009, 12:29 AM
Opalg
The example 66/66 = 1 shows that you're allowed to put two sixes together to make 66. So maybe you're also allowed to use a decimal point for .6 = 6/10. Then you could have $9 = \frac6{.6} - \frac66$.
• August 12th 2009, 02:05 AM
bruxism
Quote:

Originally Posted by Opalg
The example 66/66 = 1 shows that you're allowed to put two sixes together to make 66. So maybe you're also allowed to use a decimal point for .6 = 6/10. Then you could have $9 = \frac6{.6} - \frac66$.

Nice work. I'm sure that must be what he's looking for. It's difficult getting all the information second hand from a 15 year old but i'm fairly sure decimals must be allowed for it to work.

Now if anyone is interested or would just like to do it for fun, the assignment actually asks that it be done not only for four 6's, but also four 7's, four 8's, and four 9's. Knock yourself out everyone.