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Inspired by @Saiid's earlier question.

Using $+,-,×,÷,$ exponents and factorials, use all the numbers $1-9$ (once and only once each) to generate $300 000$.
Don't compose numbers from their digits, just operate on them.

Bonus points for keeping digits in order.
Bonus points for using all operators.

Will accept answer that I find most interesting after 24 hours.

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  • 1
    $\begingroup$ All numbers only once? $\endgroup$ – Sid Oct 4 '16 at 17:02
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I'm just a guest for now but felt like showing my answer with all bonuses.

$(1+2)((3!+4)^{\frac{5}{(6-7)(8-9)}})$
$3×10^{5/1}$
$3$e$5$
$300000$

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  • $\begingroup$ Gorgeous simplicity; well done! $\endgroup$ – Jonathan Allan Oct 7 '16 at 5:30
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Time to get 300,000, now that the body matches the title. I tried to get something fancy like my original 30,000 answer but gave up after a while:

$(3×(4+6)^5 + 7 + 9 - (8×2))/1!$

It ticks all the boxes, but in a pretty soulless way I'll admit.

BONUS: My previous answer to get 30,000

Here goes:

$(((9^5) + (3^6) + (8×7×4)) / 2) - 1 = 30000$

(I can't kill spoilers on more than the first line for some reason so I edited to omit the breakdown)

EDIT: Thanks to Jonathan for teaching me how to spoiler better so I could put in the breakdown

$((59049 + 729 + 224)/2) - 1$
$(60002/2) -1$
$30001 -1$

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  • $\begingroup$ I just noticed you can make this use all operators very easily using the fact that $1!=1$ (or even $2!=2$). $\endgroup$ – Jonathan Allan Oct 4 '16 at 17:37
  • $\begingroup$ Oh, and for multiline spoilers end your lines with two spaces. $\endgroup$ – Jonathan Allan Oct 4 '16 at 17:49
  • $\begingroup$ @Jonathan Allan, the degenerate cases of ÷1, ×1, 1! and 2! I deem less interesting ;) $\endgroup$ – Ax. Oct 4 '16 at 18:31
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$300,000$ is such a

nice round number in decimal that this is particularly easy to do with $1-9$ in order:
$(1+2+3+4)^5\times(6(7-8)+9)$

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$300,000$ can be reached using

two powers, and with digits in order, by $\dfrac{1(2^3)4(5^6)}{7+8}\times 9$

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  • $\begingroup$ Second runner up for being so darn cute. $\endgroup$ – Ax. Oct 6 '16 at 20:14
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Transforming my 30,000 answer for 300,000 that also uses all operators:

$(6!+(7+8)\times(\frac93-1))\times (4\times 5)^2$

$=((6\times 5\times 4\times 3\times 2)+15\times 2)\times 20^2$
$=(720+30)\times 400$
$=750\times 400$
$=300000$

Here is an answer (to the original 30,000) using all operators:

$(6!+\frac{8\times 4 + 7 - 9}{1})\times 2^3\times 5$

$=((6\times 5\times 4\times 3\times 2)+\frac{32 + 7 - 9}{1})\times 8\times 5$
$=(720+30)\times 40$
$=750\times 40$
$=30000$

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  • $\begingroup$ First runner up for having the largest n! $\endgroup$ – Ax. Oct 6 '16 at 20:12
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Here goes:

This is the answer to number 30000

$((8+2)^4)\times3+(9+5)-(1+6+7)$

After op has changed their number to 300000

$((9+1)^5)\times3+(4-\frac{6}{2})-(8-7)! $

Motive:

$(100000\times3)+1-1$

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In case anyone's still interested in the earlier target of $30,000$, I found an expression that seems more interesting than mine for $300,000$.

$30000=3\times 10^4=3\times (2\times 5)^4=5^4\times (2^4\times 3)=625\times 48$ so

$30000=1(2+3)^4(5(6+7)-8-9)$

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My original solution when forming the question:

$9!+((1+6^2)-3×7×8)×4×5!$

Which I thought a bit cute.

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$(1^2)+(3*4^5)-6-7-8-9-1-2-3-4-5-6-7-8-9+(1*2)$

That's my answer.

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  • $\begingroup$ "use all the numbers 1−9 (once and only once each)" $\endgroup$ – Jonathan Allan Oct 4 '16 at 18:48
  • $\begingroup$ If I did mental math right, this makes 952. Even if I made a mistake, I'm pretty sure it's not 300000. $\endgroup$ – GentlePurpleRain Oct 4 '16 at 19:02
  • $\begingroup$ It's actually 3,000. But as you said not 300,000 and too many numbers. $\endgroup$ – dcfyj Oct 4 '16 at 19:23
  • $\begingroup$ Ops, I got wrong, i'll change soon. i'm sorry $\endgroup$ – Blind Oct 6 '16 at 17:55

protected by Deusovi Oct 5 '16 at 3:21

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