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Use $2, 0, 2, 3$ and some mathematical operators to express numbers from $1$ to $100$.

Operators: $+$, $-$, $\times $, $÷$, ( ), √ (square root), $^$ (power), $!$ (factorial), $!!$ (double factorial), $P^{n}_r$ (permutation) , $C^{n}_r$ (combination)

Note: exactly two 2's, one 0 and one 3 can be used.

I found 99 of them, I'm left with 87. Is there any solution to it? If yes, what is it?

My work:

$$1=2+2-3+0$$ $$2=3!-2-2+0$$ $$3=3+2-2+0$$ $$4=3×2-2+0$$ $$5=3+2+2×0$$ $$6=3×2+2×0$$ $$7=3+2+2+0$$ $$8=3×2+2+0$$ $$9=3×(2+2-0!)$$ $$10=(3+2)×2+0$$ $$11=(3+2)×2+0!$$ $$12=3×2×2+0$$ $$13=\frac{20}2+3$$ $$14=(3\times 2+0!)\times 2$$ $$15=(2+2+0!)\times 3$$ $$16=\frac{32}2+0$$ $$17=\frac{32}2+0!$$ $$18=3^{2}\times 2+0$$ $$19=3^{2}\times 2+0!$$ $$20=23-2-0!$$ $$21=23-2+0$$ $$22=22+0\times 3$$ $$23=23+0\times 2$$ $$24=(2+2)!+0\times 3$$ $$25=20+2+3$$ $$26=23+2+0!$$ $$27=32-\frac{0!}{.2}$$ $$28=20+3!+2$$ $$29=32-2-0!$$ $$30=30+2-2$$ $$31=32-2+0!$$ $$32=32+2\times0$$ $$33=32+2-0!$$ $$34=32+2+0$$ $$35=32+2+0!$$ $$36=(3!)^{2}+2\times0$$ $$37=20\times 2-3$$ $$38=\left(\frac{0!}{.2}\right)!!+23$$ $$39={3!+2\over .2}-0!$$ $$40={3!+2\over .2}+0$$ $$41={3!+2\over .2}+0!$$ $$42=(3!)^{2}+(2+0!)!$$ $$43=20\times 2+3$$ $$44=22\times(3-0!)$$ $$45=23\times2-0!$$ $$46=23\times2+0$$ $$47=23\times2+0!$$ $$48=(3!)!!+2\times2\times0$$ $$49=(3!)!!+2-2+0!$$ $$50=(3!)!!+2+2\times0$$ $$51=(3!)!!+2+2-0!$$ $$52=30+22$$ $$53=(3!)!!+2+2+0!$$ $$54=(2+0!)^{3}\times2$$ $$55=P^{(3!+2)}_{2}-0!$$ $$56=(3!)^{2}+20$$ $$57=P^{(3!+2)}_{2}+0!$$ $$58=20\times 3-2$$ $$59=\frac{(3+2)!}{2}-0!$$ $$60=\frac{(3+2)!}{2}+0$$ $$61=\frac{(3+2)!}{2}+0!$$ $$62=20\times3+2$$ $$63=32\times2-0!$$ $$64=32\times2+0$$ $$65=22\times3-0!$$ $$66=22\times3+0$$ $$67=22\times3+0!$$ $$68=(3\times2)!!+20$$ $$69=2^{(3!)}+\frac{0!}{.2}$$ $$70=2^{(3!)}+(2+0!)!$$ $$71=(3!)!!+22+0!$$ $$72=(3!)!!+(2+2)!+0$$ $$73=(3!)!!+(2+2)!+0!$$ $$74=\frac{\frac{3}{.2}}{.2}-0!$$ $$75=\frac{\frac{3}{.2}}{.2}+0$$ $$76=\frac{\frac{3}{.2}}{.2}+0!$$ $$77=(.2)^{(-3)}-[(2+0!)!]!!$$ $$78=\frac{\left(\frac{0!}{.2}\right)!!}{.2}+3$$ $$79=\frac{(2+2)!}{.3}-0!$$ $$80=20\times(3!-2)$$ $$81=3^{(2+2)}+0$$ $$82=3^{(2+2)}+0!$$ $$83=(3!+0!)!!-22$$ $$84=2^{(3!)}+20$$ $$85=\frac{(3!-0!)!!+2}{.2}$$ $$86=\left[(3!)!!-\frac{0!}{.2}\right]\times2$$ $$88=\left(\frac{0!}{.2}\right)!!\times3!-2$$ $$89=\left[(\sqrt{.2})^{(-2)}\right]!!\times3!-0!$$ $$90=\frac{2+0!}{.2}\times3!$$ $$91=C^{(\frac{3}{.2}-0!)}_{2}$$ $$92=\left(\frac{0!}{.2}\right)!!\times3!+2$$ $$93=(3!)!!\times2-2-0!$$ $$94=(3!)!!\times2-2+0$$ $$95=(3!)!!\times2-2+0!$$ $$96=(3!)!!\times2+2\times0$$ $$97=\frac{2}{.02}-3$$ $$98=(3!)!!\times2+2+0$$ $$99=(3!)!!\times2+2+0!$$ $$100=\frac{\frac{3+0!}{.2}}{.2}$$

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    I don’t think that this question is suitable for this site, as absolutely no mathematical intuition or logic is involved in answering this. This is a puzzle, not a mathematical problem. (-1) – insipidintegrator Jun 02 '22 at 14:17
  • @insipidintegrator I agree that the question isn’t suited to this website but OP clearly said “ Note: exactly two 2's, one 0 and one 3 can be used.” – Heisenberg Jun 02 '22 at 14:20
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    Ah,my bad. I’ll edit it. – insipidintegrator Jun 02 '22 at 14:22
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    Although the expressions are a bit strange, the exercise is at last interesting. Therefore I decided to upvote it and I am curious whether somewhat finds a solution. – Peter Jun 02 '22 at 14:48
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    I appriciate your 99 works... If 87 is impossible, its proof would require a thorough search as in a similar topic https://math.stackexchange.com/questions/92230/proving-you-cant-make-2011-out-of-1-2-3-4-nice-twist-on-the-usual?rq=1 (BTW, You should add the dot operator to your list.) – aerile Jun 02 '22 at 15:10
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    $$(3!)!!\times2-2+0!=87$$ (in base 11) – emacs drives me nuts Jun 02 '22 at 15:13

0 Answers0