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You have two numbers, 9 and 20. Can you make 6?

You must use each of the two numbers exactly once.

You can use math operations: +, -, x, /, exponents, roots or factorials.

You can combine (concatenate) original numbers: e.g. 9 & 20 make 920 or 20 & 9 make 209

You cannot combine any numbers besides the original ones, split any number into separate digits,

turn numbers upside down, approximate, truncate, use summations or any math operations beyond the ones stipulated above.

Have fun!

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Bit confused about the must use each number exactly once because if you put them together to make 209 surely thats all you can do? Anyway heres one way of doing it using them more then once.

((9/20*20/9)+(9/20*20/9)+(9/20*20/9)+(9/20*20/9)+(9/20*20/9)+(9/20*20/9)) = 6

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My intention was that you get to use only the two given numbers.

Otherwise it would be trivial: I could pull a 5 (or a five) of my own out of a hat and come up with 20 - 9 - 5 = 6.

So you can't use a 5 or a 0, or a 40 (even if you spell it out). :unsure:

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Woah, that's cool

202020 = 2*3*5*7*13*37

Using 8 of the numbers:

(20*9 - 99 - 9 - 9 -9)/9

Using 7:

log(9+9+9+9) base (20 - 9 - 9)

7 again:

(920 + 9*20 - 20)/9/20

7 again!

(20-9-9)*(log(9*9*9) base 9)

Can't get it less than 7 yet

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You read my mind! That's the one I was thinking of too. Maybe there are others -- I don't know. :thumbsup:

Your intention might be to allow square-roots but you didn't mention that in the original questions. You simply said roots. And you denied some of the answers by pointing out that they used fortieths root and hence used the not-given number forty. Similarly can't I argue that you used twoth root and hence used the not-given number 2?

I know you could have always escaped this problem by using the term "square-root", rather than just "root" in your OP.

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Your intention might be to allow square-roots but you didn't mention that in the original questions. You simply said roots. And you denied some of the answers by pointing out that they used fortieths root and hence used the not-given number forty. Similarly can't I argue that you used twoth root and hence used the not-given number 2?

I know you could have always escaped this problem by using the term "square-root", rather than just "root" in your OP.

I guess I've never put a 2 in the little crook of the root symbol for square root -- to me, no number there meant the default = square root. Hmmmm.

I didn't actually want to restrict it to square roots. When I write the cube root of 8, I would use two numbers -- an 8 inside and a 3 in the crook. Are these not standard forms?

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I guess I've never put a 2 in the little crook of the root symbol for square root -- to me, no number there meant the default = square root. Hmmmm.

I didn't actually want to restrict it to square roots. When I write the cube root of 8, I would use two numbers -- an 8 inside and a 3 in the crook. Are these not standard forms?

The only roots we can use are the 9th root of x and the 20th root of x, unless of course you make another number first (do 20-9 and now you can take the 11th root of something)

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Somewhat mindless ways of getting it down to 6 numbers used:

(9*20-20-20-20)/20

(9*9-9-9-9)/9

And 5 used!

920/20-20-20

It's getting there. Gotta get it down to only one 9 and one 20 now.

:)

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