Six 10's and Endless Possibilities

clairejay

New member
Joined
Oct 22, 2012
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1
I am to use 6 10's and any operation (+,-,*,/,! etc) to come up with numbers zero through twenty. Basically, I have to create equations that equal the numbers zero through twenty ONLY using the number ten six times.
Example:
1 = (10/10) + (10/10) - (10/10)
4 = (10/10) + (10/10) + log(10) + log(10)

I need numbers 5, 7, 8, 13, 15, 16, 17, and 18. Please help!
 
I am to use 6 10's and any operation (+,-,*,/,! etc) to come up with numbers zero through twenty. Basically, I have to create equations that equal the numbers zero through twenty ONLY using the number ten six times.
Example:
1 = (10/10) + (10/10) - (10/10)
4 = (10/10) + (10/10) + log(10) + log(10)

I need numbers 5, 7, 8, 13, 15, 16, 17, and 18. Please help!

5 = (10 + 10 + 10 + 10 + 10)/10
 
I am to use 6 10's and any operation (+,-,*,/,! etc) to come up with numbers zero through twenty. Basically, I have to create equations that equal the numbers zero through twenty ONLY using the number ten six times.
Example:
1 = (10/10) + (10/10) - (10/10)
4 = (10/10) + (10/10) + log(10) + log(10)

I need numbers 5, 7, 8, 13, 15, 16, 17, and 18. Please help!
Your examples seem to show that you have ingenuity. All that is being asked of you is to use it.

\(\displaystyle 10 + (10 - 10) + log(10 * 10 * 10) = 10 + 0 + 3 = 13.\)

\(\displaystyle 10 + log(10 * 10 * 10 * 10) - log(10) = 10 + 4 - 1 = 13.\)

\(\displaystyle 10 + log(10 * 10 * 10) + log\left(\dfrac{10}{10}\right) = 10 + 3 + log(1) = 13 + 0 = 13.\)

\(\displaystyle \dfrac{10 * 10}{10} + log(10 * 10 * 10) = 10 + 3 = 13.\)

\(\displaystyle \dfrac{10 * 10 + 10}{10} + log(10 * 10) = \dfrac{110}{10} + 2 = 11 + 2 = 13.\)

I suspect that there are more, but I am lazy.
 

. \(\displaystyle 7\;=\;10 - \frac{10+10}{10} - \frac{10}{10}\)

. \(\displaystyle 8 \;=\;10 - \frac{10+10}{10}\!\cdot\!\frac{10}{10}\)

\(\displaystyle 13 \;=\;10 + \frac{10+10}{10} + \frac{10}{10}\)

\(\displaystyle 15 \;=\;10 + 10 - \frac{10 \cdot 10}{10+10}\)
 
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