How does one get a ten (10) using three 3 only. You can use any mathematical operation.?

Question

You can only use *three* 3s.

Answer 1

(3^2) + (3/3) = 10
33 - 23 = 10
33/3 -1 =10

There are infinite such examples.. but if only 3s have to be used, I think the 10 on the right side is not decimal 10, but binary or some other base

In that case, consider binary form of 3 in 3 bits
3 base 10 = 011 base 2
!(3 base 10) = !(011 base 2) =100 base 2= 4 base 10
3 base 10 + 3 base 10 - (!(3 base 10)) = 2 base 10
= 10 base 2

Hence the solution is

3 + 3 +( !(3) ) = 10

(or)

if 10 on the other side is base 9,

Then,
(3+3+3) base 10 = 9 base 10 =10 base 9
(3^3) / 3 = 9 base 10 = 10 base 9

(or)

if 10 is in base 4

then 3 + (3/3) = 10 base 4

Thus, many such examples can be cited. Just arrive at different numbers using three 3s, then take the right hand side to be to that numbers' base.

Answer 2

3+3+3+3/3=10

Answer 3

10 = 3*3+3/3

Answer 4

3 + 3 + 3 +(3*3)/3 =10

Answer 5

3 raised to the power 0(=1)+(3 multiplied by 3)
1+9=10

Answer 6

3+3+3 = 10 base 9
3*(3+inv(3)) if you recognise inverse as a mathematical operator without an implicit 1.

Answer 7

three 3's are difficult. i think you got the question wrong. if it is four 3's, then the answer should be
3 X 3 + (3 / 3) * / = divided*

how about this? how to make 100 with 4 nines?

Answer 8

log 3 (to base3) + (3x3)
= 1+9 = 10

Answer 9

Here's my solution

3 · 3 + Γ(Γ(3)) = 10

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PROOF
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Where Γ(x) is the "gamma function", whose value is
Γ(x) = (x - 1)!

Thus,
Γ(3) = (3 - 1)!


Now in the solution, we rewrite the gammas into factorials
3 · 3 + Γ(Γ(3)) = 3 · 3 + ((3 - 1)! - 1)!

Thus,
3 · 3 + Γ(Γ(3)) = 9 + (2! - 1)!

and
3 · 3 + Γ(Γ(3)) = 9 + (2 - 1)!

and
3 · 3 + Γ(Γ(3)) = 9 + 1!

and
3 · 3 + Γ(Γ(3)) = 9 + 1

and therefore,
3 · 3 + Γ(Γ(3)) = 10

QED.

^_^