12
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Use $0, 1, 2, 3, 4$ to make the any of these numbers:

$$331, 333, 435, 452, 455, 458, 461, 469, 470$$

  1. You must use all $5$ digits $0, 1, 2, 3, 4$ each exactly once. You can make multi-digit numbers out of the numbers, e.g. $120$ or $42$.

  2. The square function may NOT be used. Nor may the cube, raise to a fourth power, or any other function that raises a number to a specific power. You may use the ^ operation if you use a digit, for example, $[(10 + 3)^2 +4]$ is acceptable because $0, 1, 2, 3, 4$ is used. However, $[(10 + 3)^2 +4+2]$ can't be used because it uses an extra $2$.

  3. The integer function may NOT be used. Nor may the round, floor, ceiling, repeating or concatenation symbol, or truncate functions.

  4. The square root, multi factorial, subfactorial and decimal point may NOT be used.

  5. $+, -, *, /, (), \text{^}, \text{and }!$ (factorial) may be used for functions. Example: factorial may be used more than once, e.g. $(3!)!=720$ is acceptable.

From the numbers $0 \text{ ~ } 500$, those $9$ numbers above are the only ones I didn't get.

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  • $\begingroup$ So you are allowed to change around the order of the numbers? $\endgroup$ – Duck Jun 18 at 18:42
  • $\begingroup$ yes correct, changing the order of the numbers is allowed. $\endgroup$ – ThomasL Jun 18 at 18:45
  • $\begingroup$ Can we use combination or permutation operators? $\endgroup$ – Voldemort's Wrath Jul 26 at 19:55
  • $\begingroup$ Combination yes, permutation no $\endgroup$ – im_so_meta_even_this_acronym Jul 26 at 20:02
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    $\begingroup$ How will you choose who to give the bounty to? It looks like a lot of people contributed $\endgroup$ – PotatoLatte Jul 26 at 20:23
5
+200
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452

Using the bracket notation for the rising factorial:

$(10^{(2)}+3)\times4=(10\cdot11+3)\times4=113\cdot4=452$

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  • $\begingroup$ Do you think it's possible to actually prove it's impossible, somehow? Maybe using some size argument? Obviously it wouldn't be easy but would it be possible? $\endgroup$ – im_so_meta_even_this_acronym Jul 28 at 15:06
  • $\begingroup$ @im_so_meta_even_this_acronym; well, there's only a finite number of possibilities... $\endgroup$ – JMP Jul 28 at 19:29
  • $\begingroup$ There aren't actually, factorial exists $\endgroup$ – im_so_meta_even_this_acronym Jul 28 at 19:30
  • $\begingroup$ @im_so_meta_even_this_acronym; but how is $4!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!$ going to be of use? $\endgroup$ – JMP Jul 28 at 19:32
  • $\begingroup$ Well, it probably won't be but seems hard to prove, what if that number minus 2 divided by 3!!!!!!!!!!!!!!!!!!! or something was useful? (This clearly does not work by considering prime factors, but you get the point). $\endgroup$ – im_so_meta_even_this_acronym Jul 28 at 19:54
8
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Partial Answer:

470

$4\times(3+2)!-10 = 470$

455 (by @ThomasL)

$(4!)^2-(3!-0!)!-1 = 455$

333 (by @ripkoops)

$213+(0!+4)! = 333$

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  • 2
    $\begingroup$ I managed to get 455: (4!)^2-(3!-0!)!-1 = 576-120-1=455 $\endgroup$ – ThomasL Jun 20 at 20:47
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    $\begingroup$ 333 = 213+(0!+4)! $\endgroup$ – ripkoops Jun 21 at 1:06
  • $\begingroup$ ThomasL, ripkoops Thanks! I've aggregated these comment answers into the answer above $\endgroup$ – Caleb Jun 25 at 16:25
4
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Interpretation of the brackets as being the binomial coefficient I was able to get four more numbers 435, 458, 461, and 469:

$435 = \binom{30}{2}\times1^4$
$458 = \binom{30}{2}+4!-1$
$461 = \binom{31}{2}-4+0$
$469 = \binom{31}{2}+4+0$

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  • $\begingroup$ two numbers are left to crack: 331 and 452 $\endgroup$ – ThomasL Jun 27 at 20:17
4
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EDIT: My original solutions were invalid, sorry for being a complete idiot.

New solution for 331:

$\dbinom{2\cdot3!-1}{4}+0!$

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  • $\begingroup$ well the OP doesn't say you can't use binomial... $\endgroup$ – JMP Jul 5 at 7:02
  • $\begingroup$ I assumed it was ok as the OP used binomial in the post directly above mines. Just checking I didn't blatantly violate one of the rules again :) $\endgroup$ – im_so_meta_even_this_acronym Jul 5 at 7:05
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    $\begingroup$ I guess so then! I'd assumed binomial's were illegal otherwise you can get away with almost anything , but if the O says they're OK, then they must be! $\endgroup$ – JMP Jul 5 at 7:12
1
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Some "almost-solutions" (with one of the digits repeated twice) for 452, the only number which remains unsolved at the time of posting (I'm posting this as ideas, because somebody may rework them):

with 2 zeroes: $452=\binom{4!/2-0!}{3!}-10$
with 2 ones: $452=\binom{4!/2-1}{3!}-10$
with 2 threes: $452=(3!+1^0)\cdot2^{3!}+4$
with 2 fours: $452=\binom{10+2}{4}-43$

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0
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Possible for 452

I'm not sure whether this is allowed or not but here goes:

$0!=1$
$1+1=2$
$3+2=5$
$concat(concat(4, 5), 2)$

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  • $\begingroup$ sorry, but concat is not allowed as already mentioned in rule 3 $\endgroup$ – ThomasL Jul 26 at 20:08
  • $\begingroup$ @ThomasL - Whoopsie! I must have missed that... $\endgroup$ – Voldemort's Wrath Jul 26 at 20:12
  • $\begingroup$ Don't worry I did this too before :P $\endgroup$ – im_so_meta_even_this_acronym Jul 26 at 20:19

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