2

Part one: You are presented with ten incomplete equations:

  • 0 0 0 = 6

  • 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

Your job is to fill in mathematical operators in order to make all equations true.

  1. You can not connect two digits into one number. So for example you cannot connect two fives into fifty five. Each digit is it's own separate number.
  2. You cannot add in more numbers, including in operators. So squaring is invalid, because in order to square you need to add in the number two. A square root is allowed, but not any other root that requires additional numbers.
  3. A square root of a square root is a fourth root, which is also not allowed
  4. You cannot change anything but the left side of the equation
  5. Functions such as sine, tangent etc. Are not allowed.
  6. Using ^ is not accepted

Part two: using the same rules, can you fill in

  • n n n n = 6
Lavigo
  • 143
  • 8
  • @boboquack I wasn't aware of this, but my version is a bit tweaked – Lavigo Oct 25 '17 at 08:16
  • @boboquack If it does count as a duplicate, do as you please – Lavigo Oct 25 '17 at 08:19
  • That comment was automatically generated by the system. My observation is that the only difference is part two - if you changed part two to be the whole question, it would be distinct enough, but at the moment most of the question is the same is the linked question. – boboquack Oct 25 '17 at 08:23
  • The rules are also different – Lavigo Oct 25 '17 at 08:40
  • All the solutions in the accepted answer work here. – boboquack Oct 25 '17 at 09:01
  • @boboquack true, but 8 - v(v(8+8) is also correct in the first one, and is not in mine. – Lavigo Oct 25 '17 at 09:15
  • The most extreme case has to be:(2^3)- v(v(2^3+2^3)) which some people have tried to suggest in real life, which is why it is a rule – Lavigo Oct 25 '17 at 09:19

2 Answers2

5

Part One:

(0! + 0! + 0!)! = 6
(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

Part Two:

((n!/n!)+(n!/n!))# = 6

# is being used as Primorial, which is the product of the first x prime numbers. In this case, the first two primes, which is 2*3=6

Alternatively:
((n! + n! + n!) / n!)! = 6

Timoris
  • 1,331
  • 10
  • 12
  • A. I said a square root of a square root is not allowed B. What about part two? – Lavigo Oct 25 '17 at 08:13
  • Try typing √a (√a) instead of va or even $\sqrt a$ ($\sqrt a$). – boboquack Oct 25 '17 at 08:13
  • second equation requires a correction as either of the below shown ways: (1) (1 + 1 + 1) ! OR (2) (1! + 1! + 1!) ! but not the given one! – Mea Culpa Nay Oct 25 '17 at 08:15
  • Thanks for spotting my slight error with the 1s, Mea. What about now, @Lavigo ? – Timoris Oct 25 '17 at 08:39
  • If a primorial is going to be considered an invalid function, then so should a factorial, (primorial is just "prime factorial" in a way, hence the name) which makes the entire question impossible. – Timoris Oct 25 '17 at 08:46
  • As far as I understand, your solution comes out as two. Am I misunderstanding anything? – Lavigo Oct 25 '17 at 08:50
  • I get it now. Well done, I did not think of that. – Lavigo Oct 25 '17 at 08:52
  • An alternative solution for part 2 to is to use the solution for part 1 and replace one of the numbers by $\sqrt{nn}$. You would need to change the part 1 answer for 9 as the substitution would lead to a root of a root, but you could use $9-\sqrt{\sqrt{9}\sqrt{9}} = 6$ for example. – Jaap Scherphuis Oct 25 '17 at 08:56
  • 1
    It took me a while to think of it. The tricky part was trying to make an equation that would work for 0 as well as all other integers. I knew I'd have to use factorials, but then I also thought that might screw with larger numbers. Though now I think about it, I think ((n!+n!+n!)/n!)! would also work. – Timoris Oct 25 '17 at 08:57
  • It doesn't say n has to be an integer, or from 0-9 (@Jaap). Also, I thought the primorial p# was the product of the primes less than or equal to p? – boboquack Oct 25 '17 at 08:59
  • @boboquack That's what I thought too, but then I looked it up – Lavigo Oct 25 '17 at 09:03
  • Aha! But $p_5=11$ - so $p_5#=11#=2310$, which is what I said. Importantly, $p_1=2$, so $p_1#=2#=2$ (@Lavigo). – boboquack Oct 25 '17 at 09:05
  • I copied the wrong bit. xD It is written in the first introduction section on Wikipedia it has two definitons and states them. I will admit I used the less-used of the two, but the puzzle did not say I was not allowed to. – Timoris Oct 25 '17 at 09:06
  • @Timoris ((n!+n!+n!)/n!)! Was the one I was going for. edit it into your answer! – Lavigo Oct 25 '17 at 09:09
  • @Timoris Very true. – Lavigo Oct 25 '17 at 09:12
  • A * and a - is much simpler for 3. – vacawama Oct 25 '17 at 12:14
3

In continuation to earlier answer, here is one for part two:

((n+n+n)/n) ! = 6 for all n >0.

And if n = 0,

((n! + n! + n!))! * n! or ((n! + n! + n!))! / n!

And here is one more, perhaps

( log n (n * n *n ) )! that is log to the base n (subscript) of n cubed.

Mea Culpa Nay
  • 8,626
  • 1
  • 13
  • 45