1
$\begingroup$

The title explains it pretty much, make $7777/6666$ equal $-212$. The following operators can be used.

$+,-,*,/,\hat{},()$

Rules:

  1. Each number can only be used once
  2. You can move the numbers around as much as you want, but the '7777'(and whatever it gets turned into) must stay on the opposite side from the '6666'(and whatever it gets turned into)
  3. Each operator can be used only once on each side of the fraction(so twice total)
  4. You can operate with only two numbers at a time.(e.g. you can't do $7-777$, you could do at most $7-7$)
  5. It is assumed that when two numbers are operated with, they are removed from the equation and then put back in. This is not proper math, but hey, it's a puzzle.(basically, if you wanted to do $7+7$ with the first two 7's, you would do $7+7=14$, then put it back in the problem, leaving you $1477$. Yes, I know it's illogical and improper math, but it's how I made the puzzle, so deal with it.)
  6. You can round, but only decimals to the nearest 1.
  7. The only exception to the 4th rule is when dividing the '7777' section(not necessarily 7777) and the '6666' section(not necessarily 6666). In that instance, you divide the whole problem, not just the middle two digits, whatever they are.

Note - () is only for grouping and cannot be used for multiplication in this problem.

Another Note - If there is a negative number in the middle, e.g. $77(-7)7$, this makes the whole thing negative, leaving $-7777$ and eliminating a possible loophole in the third rule.

As an example of the process, this is a sample solution for the equation $8765/4321=\frac15$.

$8\ 7\ 6\ 5\ /\ 4\ 3\ 2\ 1\ =\ \frac15$

$(8\ *\ 7)\ 6\ 5\ /\ (4\ /\ 2)\ 3\ 1\ =\ \frac15$ // Rule 2: you can move the numbers however you want

$`5\ (6`\ -\ 6)\ 5\ /\ (`2`\ * \ 3)\ 1\ =\ \frac15$ // Rule 5: the numbers get inserted back into the puzzle after operations

$(5\ +\ `0`)\ 5\ /\ (`6`\ -\ 1)\ =\ \frac15$

$(`5`\ /\ 5)\ /\ `5`\ =\ \frac15$

$`1`\ /\ 5\ =\ \frac15$

†numbers surrounded by $` `$ are the results of the previous operation

Is there a reason you guys are downvoting? If there is, could you post a comment about why? I know you have enough reputation to comment if you have enough to downvote.

Improve this question
$\endgroup$
13
  • $\begingroup$ Can we change -212 in any way? $\endgroup$
    – Beastly Gerbil
    Oct 4, 2016 at 15:07
  • $\begingroup$ @BeastlyGerbil nope, the target number cannot be changed in any way. $\endgroup$
    – TrojanByAccident
    Oct 4, 2016 at 15:11
  • $\begingroup$ I absolutely do not understand this puzzle. (Especially rules 5, 7, and the another note at the end) - and how is the answer by dcfyj correct? What does "7 7 (7 + 7)" mean? $\endgroup$
    – Nova
    Oct 4, 2016 at 17:44
  • 1
    $\begingroup$ As far as I understand now, this puzzle is not about making a correct equation like any other, it's about creating a line of numbers where you can add operators on the fly to get the correct result. So you take the original equation and then always chose two numbers, calculate an operator on them and then insert the result again in the equation. $\endgroup$
    – Nova
    Oct 4, 2016 at 18:02
  • 1
    $\begingroup$ @DavidStarkey done $\endgroup$
    – TrojanByAccident
    Oct 4, 2016 at 20:49

2 Answers 2

Reset to default
9
$\begingroup$

Using Sconibulus's idea for the 7s here's my solution:

$7\ 7\ (7+7) / (6*6)\ 6\ 6 = -212 \\ 7\ (7-1)\ 4 / (3-6)\ 6\ 6 = -212 \\ (7*6)\ 4 / -3\ (6/6) = -212 \\ 4\ 2\ 4 / (-3+1) = -212 \\ 4\ 2\ 4 / -2 = -212$

Improve this answer
$\endgroup$
2
  • $\begingroup$ @gtwebb it does... $\endgroup$
    – TrojanByAccident
    Oct 4, 2016 at 16:22
  • $\begingroup$ I'll add () for clarification $\endgroup$
    – dcfyj
    Oct 4, 2016 at 16:25
6
$\begingroup$

Here's an answer that I'm pretty sure works:

$(7*(7-(7+7)))/(6/6-(6+6))=-212$
$(7*(7-14))/(1-12) = -212$
$(7*64)/(-02) = -212$ //This step declared illegal because apparently $-0 \neq 0$ :)
$424/-2= -212$

Improve this answer
$\endgroup$
1
  • $\begingroup$ actually, 1-12=2, since 1-1=0, then add the 2 on the end, so 1-12=02 $\endgroup$
    – TrojanByAccident
    Oct 4, 2016 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.