Get $5$ by doing any operations with four $7$s

Question

How can one combine four sevens with elementary operations to get $5$? For example $$\dfrac{(7+7)\times7}{7}$$ (though that does not equal $5$). I am not able to do this. Can you solve it or prove that it's impossible?

Answer 1

How about:

$$7 - \frac{7+7}{7} = 5$$

$$7 - \log_7 (7·7) = 5$$

$$7 - \frac{\ln (7·7)}{\ln 7} = 5$$

$$\left\lfloor \sqrt{\frac{7^7}{7!}} - 7\right\rfloor = 5$$

$$\lfloor 7\sin 777^\circ\rfloor = 5$$

$$\lfloor 7\cos 7^\circ\rfloor - \frac{7}{7} = 5$$

$$\lfloor 7\cos 7\rfloor = 5 \text{ using radians}$$

You can also use base $174$ and write:

$$\sqrt{\frac{77}{7·7}} = 5$$

That can also reduce the amount of sevens by one if you write:

$$\frac{\sqrt{77}}{7} = 5$$

Answer 2

Perhaps: $$7-\frac{7+7}{7}=7-2=5$$

Answer 3

Maybe:

$ \frac{((7^2-7)-7)}7=5$

Answer 4

Don't forget $$\frac{7! \mod 77}{7}$$ (and I haven't used any sneaky $2$s, either).

To get it down to three $7$s, you might try $$7 - \left\lceil .7 + .7 \right\rceil$$

You can also do it with only two $7$s: $$\left\lfloor{\sqrt 7 + \sqrt 7}\right\rfloor$$

(Can anyone do it with a single $7$?)

Answer 5

7-(7+7)/7

I started looking for a way to use the modulus operator, but this is too simple.

Answer 6

How about

$$ f(7, 7, 7, 7) $$

where we define

$$ f(x, y, z, w) = 5 $$

for all $x, y, z, w$. (Clearly, the constant function on four parameters is an elementary operation.)

Answer 7

How about $$ \lfloor7-\ln7\rfloor=5 $$ I use only two $7$s.

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$$ \lfloor7+7-\ln 7!\rfloor=5 $$ That's three $7$s.

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But if you insist on using four $7$s $$ \lfloor7-7+7-\ln 7\rfloor=5. $$

Answer 8

With four 7's you can get any positive integer you want by just changing the number of squar roots in the following equation:

$$\frac{\ln\bigg{(}\frac{\ln(7)}{\ln(\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{7}}}}})}\bigg{)}}{\ln\bigg{(}\frac{\ln(7)}{\ln(\sqrt{7})}\bigg{)}}=5$$

For example, you could get 35 by using 35 square roots. We aren't even really using the 7, you could equally use any other integer >1.

EDIT: Moved to only four 7's instead of five, based on Darth's suggestion.

Answer 9

OK, here's the ultimate: you can do it with a single $7$. It starts with the observation that when $n$ is smallish, $\ln n!$ is a little, but not much, greater than $n$. By repeatedly applying factorial, natural log and floor, we can get the value to creep up to around $5^2$.

$$\left\lfloor\sqrt{\ln\left(\left\lfloor\ln \left(\left\lfloor\ln \left(\left\lfloor\ln \left(7!\right)\right\rfloor!\right)\right\rfloor!\right)\right\rfloor!\right)}\right\rfloor = 5$$

I know you had four $7$s to use. But think of all the things you can do with the three $7$s you'll have left over!

Answer 10

How about : $\Delta_{Q(\sqrt7)}/7+7/7$, $\Delta$ being the discriminant of a number field.