# A maths home work quiz

I give you 3 digits and a result and you must put all the signs necessary to restore the equality.

I’ll give you an example. The remainder you solve by yourself.

2 + 2 + 2 = 6

Easy isn’t this? It’s the same for the remainder.

1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6

## 0 thoughts on “A maths home work quiz”

1. Anonymous says:

Cool and well done. Not the only solutions, so I will post all the answers sent in tomorrow. Cheers

2. Cool and well done. Not the only solutions, so I will post all the answers sent in tomorrow. Cheers

3. Jan Pompe says:

I’ll wait for the solutions I have one for all except the first and am going to give up on it.

4. Jan Pompe says:

I’ll wait for the solutions I have one for all except the first and am going to give up on it.

5. Carl says:

I think the solutions to 3,4,5,6,7,8 and 9 are fairly straightforward. However, number 1 isn’t.
I’ve tried to reproduce it below, the right hand side should read “the infinite root of 6”.

1 x 1 x 1 = ∞√6

6. I think the solutions to 3,4,5,6,7,8 and 9 are fairly straightforward. However, number 1 isn’t.
I’ve tried to reproduce it below, the right hand side should read “the infinite root of 6”.

1 x 1 x 1 = âˆžâˆš6

7. Jan Pompe says:

carl

1*1*1 = 6^0 did cross the mind but I thought it might not be legit.

8. Jan Pompe says:

carl

1*1*1 = 6^0 did cross the mind but I thought it might not be legit.

9. φ(n) (Euler’s totient function) is your friend…

10. Ï†(n) (Euler’s totient function) is your friend…

11. Bob D says:

Are other digits allowed as subscripts or superscripts?

12. Bob D says:

Are other digits allowed as subscripts or superscripts?

13. Anonymous says:

Bob D: Yes.
Jan: 6^0 Works for me

14. Jan Pompe says:

Thanks David,

for both the fun and the answer.

I wasn’t sure whether we we could operate on both sides.

Now I’m really curious about what Josh might be thinking. Josh?

15. Jan Pompe says:

Thanks David,

for both the fun and the answer.

I wasn’t sure whether we we could operate on both sides.

Now I’m really curious about what Josh might be thinking. Josh?

16. Anonymous says:

φ(3+3+3) =6
φ(6+6+6) =6

17. Ï†(3+3+3) =6
Ï†(6+6+6) =6

18. Jan Pompe says:

David

Ahah.

I am are others?

19. Jan Pompe says:

David

Ahah.

I am are others?

20. Anonymous says:

OK: Where n is any number:

(n^0 + n^0 + n^0)! = 6

I’ll collate all the answers into a post if you want to surrender them now.

David Stockwells last blog post..A maths home work quiz

21. OK: Where n is any number:

(n^0 + n^0 + n^0)! = 6

I’ll collate all the answers into a post if you want to surrender them now.

David Stockwells last blog post..A maths home work quiz

22. 1! × ⌊tan 1⌋ + log 1 = ⌈cos 6⌉

23. 1! Ã— âŒŠtan 1âŒ‹ + log 1 = âŒˆcos 6âŒ‰

â˜º

24. Geoff Sherrington says:

Good mathematicians start conversations either by drawing a triple integral first, or by explaining “On one hand….”

Digital solutions:

On one hand I found 5 fingers + 1 thumb = 6.

On the other hand, I found 4 fingers + 1 thumb = 5.

(Noted asymmetry and mirroring).

While looking for a nice solution in hand, I deduced that people smart enough to know about Euler’s Totient Function are w…….s. (wiki readers).

Time for solution = 0.3 milliseconds.

(This has not been peer reviewed).

25. Geoff Sherrington says:

Good mathematicians start conversations either by drawing a triple integral first, or by explaining “On one hand….”

Digital solutions:

On one hand I found 5 fingers + 1 thumb = 6.

On the other hand, I found 4 fingers + 1 thumb = 5.

(Noted asymmetry and mirroring).

While looking for a nice solution in hand, I deduced that people smart enough to know about Eulerâ€™s Totient Function are w…….s. (wiki readers).

Time for solution = 0.3 milliseconds.

(This has not been peer reviewed).

26. Ian Random says:

(1+1+1)!=6

27. Ian Random says:

(1+1+1)!=6

28. Geoff Sherrington says:

More serious this time, does this fit the rules of the game? Might not work for n = 0.

{[(n)! – (n-1)!]/3 + [(n)! – (n-1)!]/3 + [(n)! – (n-1)!]/3 } = n

I have never really thought about the sign of the factorial of a negative integer. I guess it alternates.

Thought time: 2 weeks

29. Geoff Sherrington says:

More serious this time, does this fit the rules of the game? Might not work for n = 0.

{[(n)! â€“ (n-1)!]/3 + [(n)! â€“ (n-1)!]/3 + [(n)! â€“ (n-1)!]/3 } = n

I have never really thought about the sign of the factorial of a negative integer. I guess it alternates.

Thought time: 2 weeks

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