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Use exactly three 7s in every expression and no other digits/numbers.

Choose from among addition, subtraction, division, and/or multiplication operations.

You may use parentheses, brackets, and/or braces for grouping and/or multiplication, as needed.

You may use decimal points. In addition to this/more specifically to this, you may use the decimal point where it immediately precedes the digit and at the same time is above the digit to indicate that digit repeats forever. For example, .1 (with a decimal point over the one) is equal to .111... = 1/9.

No concatenation is allowed.

You may not use factorial signs.

You may use a finite number of square roots.

Exponentiation is not allowed.

No other characters or operations may be used.

Note: Expressions with forms of A + B and B + A will be considered the same, as will be -A + B with B - A, and A(B) with B(A), and A(B) + C with C + A(B), etc.

Create a minimum of three essentially different solutions based on the note above.

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  • $\begingroup$ A remedy for working around Latex for showing .7 with a decimal point over the 7 might be to write r(.7) to indicate the repeating decimal with the digit of 7 or something similar to that. $\endgroup$
    – Olive Stemforn
    Feb 29, 2020 at 17:45
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    $\begingroup$ I have a very low opinion of people changing the question to exclude an otherwise valid answer. -1. $\endgroup$
    – David G.
    Mar 1, 2020 at 3:37
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    $\begingroup$ Why did you only limit specifically square roots to be used finitely often? It is as if you made the extra rule to remove an answer. You may only use a finite number of operands could have been an extension without headhunting. Also, why not just appreciate people's imagination? $\endgroup$
    – Therkel
    Mar 1, 2020 at 8:38
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    $\begingroup$ Please keep comments civil, and do not use flags to escalate a difference of opinion. Some comments here have been swept up. I'll note that edits to a question that invalidate existing answers are frowned upon here and that inventive abuse of rules loopholes in formation-of-numbers puzzles is a bit of a tradition here, so huffily asserting that a particular unintended use of an allowed operation is "bogus" seems unwarranted. $\endgroup$
    – Rubio ♦
    Mar 1, 2020 at 8:44

6 Answers 6

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We can get David G.'s general idea by just moving things around a bit. I included the others as well for completeness.

$\frac{\sqrt{.\dot7} + \sqrt{7}}{\sqrt{.\dot7}} = \frac{\sqrt{.\dot7}}{\sqrt{.\dot7}} + \frac{\sqrt{7}}{\frac{\sqrt{7}}{\sqrt{9}}} = 1 + 3 = 4$

$\sqrt{7 + (7 / .\dot7)} $


$ 7 - \sqrt{7 / .\dot7} $

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  • $\begingroup$ @ hdsdv - You have a correct third correct solution out of three solutions. $\endgroup$
    – Olive Stemforn
    Mar 1, 2020 at 4:18
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I've got a third answer.

By defining $f_0(x) = x$ and $f_n(x) = \sqrt{f_{n-1}(x)}$, we find that $\displaystyle \lim_{x\to\infty}{f_{n}(7)} = 1$. In short $f_n(x)$ is n square roots if x. Referring to the limit as $f_\infty(7)$ the third solution is:

$f_\infty(7) + \sqrt{7 / .\dot7}$

Note that this notation is only because I can't write infinitely many $\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{...}}}}}$

Since @Olive objects to the previous notation, an alternate way of writing this might be
$\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{...\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{7}}}}}}}}}} + \sqrt{7 / .\dot7}$

The other two, as previously described by others, are:

$\sqrt{7 + (7 / .\dot7)} $

and

$ 7 - \sqrt{7 / .\dot7} $

I will admit to being disappointed not to find a solution involving

.7

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  • $\begingroup$ @ David G. - "No other characters or operations may be used." It is not valid. $\endgroup$
    – Olive Stemforn
    Mar 1, 2020 at 1:19
  • $\begingroup$ @OliveStemforn The operation is an infinite number of square roots. Recommend an alternate rendering. $\endgroup$
    – David G.
    Mar 1, 2020 at 2:33
  • $\begingroup$ @ David G. - There is an edit in the instructions now regarding limitations for the number of square roots. $\endgroup$
    – Olive Stemforn
    Mar 1, 2020 at 3:04
  • $\begingroup$ @OliveStemforn post-changing the rules of the question in order to erase unwanted answers is a very stupid thing to do. $\endgroup$
    – Quotenbanane
    Mar 1, 2020 at 23:26
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I think the answer is:

$\sqrt{(7/.\dot7)+7}$

Explanation:

$$7/.\dot7 = 9$$ $$9 + 7 = 16$$ $$\sqrt{16} = 4$$

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    $\begingroup$ @ shoover - I made an edit to insert required decimal points in your solution. This pertains to the fourth line down in the instructions of the puzzle. You have one of the three correct solutions so far. $\endgroup$
    – Olive Stemforn
    Feb 29, 2020 at 18:35
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One answer:

$7 - \sqrt{7/.\dot7}$, where $.\dot7 = 7/9$

Previous answer from shoover:

$\sqrt{(7/.\dot7)+7}$

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  • $\begingroup$ @ JS1 - You have the second of three solutions I am seeking so far. $\endgroup$
    – Olive Stemforn
    Feb 29, 2020 at 21:44
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One of the answers is:

$\sqrt{(7/.\dot7)+7}$

It is not valid without the decimal point (as shown here) in front of the 7 that is in the denominator.

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Three answers:
1 .

Using $4 = 7 - 3$:
$4 = 7 - \sqrt{\frac{7}{.\dot{7}}}$

2 .

Using $4 = \sqrt{16}$:
$4 = \sqrt{7 + \frac{7}{.\dot{7}}}$

3 .

I might've broken the rules a little but, using $4 \approx \sqrt{\sqrt{63}}$:
$4 = \sqrt{\sqrt{7 \div \frac{.\dot{7}}{7}}}$

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