Table of contents for Quizzes
- A maths home work quiz
- The home work quiz
OK, here are the solutions I received to the home work quiz. Thanks to Alan D. McIntire, Peter Gallagher, Jan Pompe, BobD and Josh for contributions.
1:
(1+1+1)!=6
1*1*1 = 6^0
1 x 1 x 1 = ∞√6
11|4 + 1 = 6 where the 4 is a subscript indicating “base 4″
1×1+1 = φ(6)
2:
(2*2) - 2 = 6
2 + 2 + 2 = 6
3:
(3*3) - 3 = 6
φ(3+3+3) =6
4:
4^(1/2) + 4^(1/2) + 4^(1/2) = 6
4 + 4 - 4^(1/2) = 6
5:
(5/5) + 5 = 6
5 + 5 / 5 = 6
6:
6+6-6=6
6 - 6 + 6 = 6
φ(6+6+6) =6
7:
7 - (7/7) = 6
8:
8-8^0-8^0 = 6
8^(1/3) + 8^(1/3) + 8^(1/3) = 6
8 - (8+8)^.25 = 6
Γ(8 -8/8) = 6!
9:
(9 + 9) / 9^ (1/2) = 6
9^(1/2) x 9^(1/2) – 9^(1/2) = 6
9 - (9 / 9^.5) = 6
φ is Euler’s Totient Function.
Γ is the Gamma function.
And my solution where n is any number:
(n^0 + n^0 + n^0)! = 6
5 responses so far ↓
just to test I’ve got this totient thingy
hows
1×1+1 = φ(6)
A couple of others:
1:
11|4 + 1 = 6 where the 4 is a subscript indicating “base 4″
4:
4 + 4 - 4^(1/2) = 6
By the way:
4! / 4 = 6
is true but lacks a 4.
I think solutions like
8 - (8+8)^.25 = 6
and
(9 + 9) / 9^ (1/2) = 6
are cheating because of those additional numbers, 0.25 and 1/2.
Why are ^0.25 and ^ 1/2
allowed, and not -21, resulting in trivial solutions
like 9+9+9 -21 =6 .
You could have gotten around it by throwing in SQRT(SQRT) and SQRT.
My own solution to 8 was
Gamma (8 -8/8) = 6!
I couldn’t solve the 9 series without the
SQRT.
Its all good. Hope people learned something. Cheers
Alan #3
there is a difference between defining operations on terms and adding extra ones.
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