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Use $0, 1, 2, 3, 4$ to make the any of those numbers:
$$331, 333, 435, 452, 455, 458, 461, 469, 470$$
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You should use all $5$ digits $0, 1, 2, 3, 4$ every precisely as soon as. You could make multi-digit numbers out of the numbers, e.g. $120$ or $42$.
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The sq. perform could NOT be used. Nor could the dice, increase to a fourth energy, or every other perform that raises a quantity to a selected energy.
You could use the ^ operation if you happen to use a digit, for instance, $[(10 + 3)^2 +4]$ is suitable as a result of $0, 1, 2, 3, 4$ is used. However, $[(10 + 3)^2 +4+2]$ cannot be used as a result of it makes use of an additional $2$. -
The integer perform could NOT be used. Nor could the spherical, flooring, ceiling, repeating or concatenation image, or truncate features.
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The sq. root, multi factorial, subfactorial and decimal level could NOT be used.
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$+, -, *, /, (), textual content{^}, textual content{and }!$ (factorial) could also be used for features. Example: factorial could also be used greater than as soon as, e.g. $(3!)!=720$ is suitable.
From the numbers $0 textual content{ ~ } 500$, these $9$ numbers above are the one ones I did not get.
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9
452
Using the bracket notation for the rising factorial:
$(10^{(2)}+3)times4=(10cdot11+3)times4=113cdot4=452$
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9
Partial Answer:
470
$4times(3+2)!-10 = 470$
455 (by @ThomasL)
$(4!)^2-(3!-0!)!-1 = 455$
333 (by @ripkoops)
$213+(0!+4)! = 333$
331 (by @ThomasL)
$((3!)!-10)/2-4!$
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5
Interpretation of the brackets as being the binomial coefficient I used to be capable of get 4 extra 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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1
Some “almost-solutions” (with one of many digits repeated twice) for 452, the one quantity which stays unsolved on the time of posting (I’m posting this as concepts, as a result of someone could 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: $452=binom{10+2}{4}-43$
answered Jul 5, 2019 at 10:50
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I’m undecided whether or not that is allowed or not however right here goes:
$0!=1$
$1+1=2$
$3+2=5$
$concat(concat(4, 5), 2)$
answered Jul 26, 2019 at 20:06
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3
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