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Create all the numbers from $1$ to $100$ using the numbers $1,$$3,$$3,$ and $7.$

You can only use each number once, except for the $3,$ of which you have two.
You can use addition $(x+y),$ subtraction $(x−y),$ division $(\frac{x}{y}),$ multiplication $(x\times y),$ exponentiation $(x^y),$ roots $(y\sqrt x)$, factorials $(x!)$ and ceiling and floor $(\lceil x\rceil,\lfloor y\rfloor)$.
You can combine numbers like $1$ and $7$ to $17$ etc.
Use of any types of brackets are also allowed.
Good luck!

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  • $\begingroup$ Hi @H_D I'll try this ;) +1 for your first post ;) $\endgroup$ – Omega Krypton Jan 29 at 11:28
  • $\begingroup$ We are not allowed to use brackets? :( $\endgroup$ – Tweakimp Jan 29 at 11:46
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    $\begingroup$ Use of brackets are allowed. Don't be sad. $\endgroup$ – H_D Jan 29 at 12:07
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    $\begingroup$ Why the downvote? This site should not be discriminating new members of our community, please check out the Code of Conduct. We should all be community-minded and not simply drown new members with -1s. This is a legitimate question, where I do not see the point of downvoting. $\endgroup$ – Omega Krypton Jan 29 at 12:40
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    $\begingroup$ I agree with both of @OmegaKrypton's comments, although adding such a restriction after there are already 3 answers is probably bad form. If we're talking restrictions, I would love to see concatenation banished from these puzzles in general. $\endgroup$ – tilper Jan 29 at 12:57
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1 number missing. I give up

$0 = 7-3-3-1$
$1 = 7 \times 1 - 3 - 3$
$2 = 7 + 1 - 3 - 3$
$3 = \frac{7 - 1 + 3}{3}$
$4 = \sqrt{17-\frac{3}{3}}$
$5 = 3 \times (3 + 1) - 7$
$6 = 7 - 1 + 3 - 3$
$7 = 7 \times 1 - 3 + 3$
$8 = 7 + 1 - 3 + 3$
$9 = 13 + 3 - 7$
$10 = 3^3 - 17$
$11 = 17 - 3 - 3$
$12 = 3 \times (7 - 3) \times 1$
$13 = 3 \times (7 - 3) + 1$
$14 = 7 \times (1 + \frac{3}{3})$
$15 = 3 \times (7 - 3 + 1)$
$16 = 17 - \frac{3}{3}$
$17 = 17 - 3 + 3$
$18 = 17 + \frac{3}{3}$
$19 = 7 \times 3 + 1 - 3$
$20 = 3^3 - 7 \times 1$
$21 = 3^3 - 7 + 1$
$22 = (7-3)! - 3 + 1$
$23 = 7 \times 3 + 3 - 1$
$24 = \frac{73 - 1}{3}$
$25 = 33 - 7 - 1$
$26 = 33 - 7 \times 1$
$27 = 33 - 7 + 1$
$28 = 3 \times 7 + 3! + 1$
$29 = 3! \times 3! - 7 \times 1$
$30 = 37 - 3! - 1$
$31 = 37 - 3! \times 1$
$32 = 37 - 3! + 1$
$33 = 37 - 3 - 1$
$34 = 37 - 3 \times 1$
$35 = 37 - 3 + 1$
$36 = (7-1) \times (3+3)$
$37 = 3! \times (3! - 1) + 7$
$38 = 3! \times 7 - 3 - 1$
$39 = 3! \times 7 - 3 \times 1$
$40 = 3! \times 7 - 3 + 1$
$41 = 37 + 3 + 1$
$42 = 37 + 3! - 1$
$43 = 37 + 3! \times 1$
$44 = 37 + 3! + 1$
$45 = 3! \times 7 + 3 \times 1$
$46 = 3! \times 7 + 3 + 1$
$47 = 3! \times 7 + 3! - 1$
$48 = 3! \times 7 + 3! \times 1$
$49 = 7 \times (3+3+1)$
$50 = (7+3) \times (3!-1)$
$51 = 17 \times (3! - 3)$
$52 = 13 \times (7-3)$
$53 = \lfloor\sqrt{7^3}\rfloor \times 3 - 1$
$54 = 17 + 3^3$
$55 = 7 \times 3 + (3+1)!$
$56 = (3\times 3 - 1) \times 7$
$57 = 17 \times 3 + 3!$
$58 = \lfloor\frac{7^3}{3!}\rfloor + 1$
$59 = 3! \times (7+3) - 1$
$60 = 3! \times (7+3) \times 1$
$61 = 3! \times (7+3) + 1$
$62 = 7\times 3^3 - 1$
$63 = 7\times 3^3 \times 1$
$64 = 7\times 3^3 + 1$
$65 = (7+3!) \times (3!-1) $
$66 = 7\times 3! + (3+1)!$
$67 = (3+1)^3 + \lceil\sqrt{7}\rceil = 64 + 3$
$68 = \lfloor\frac{7^3}{3!-1}\rfloor$
$69 = \lceil\frac{7^3}{3!-1}\rceil$
$70 = (7 + 3) \times (3! + 1)$
$71 = 71 -3 + 3$
$72 = (3! + 3!) \times (7-1)$
$73 = 73 + \lfloor\frac{1}{3}\rfloor$
$74 = 3^{3+1} - 7$
$75 = 73 + 3 - 1$
$76 = 73 + 3 \times 1$
$77 = 73 + 3 + 1$
$78 = 3! \times (7 + 3!) \times 1$
$79 = 3! \times (7 + 3!) + 1$
$80 = 3^{7-3} - 1$
$81 = 3^{7-3} \times 1$
$82 = 3^{7-3} + 1$
$83 = 7 \times (3! + 3!) - 1$
$84 = 7 \times (3! + 3!) \times 1$
$85 = 7 \times (3! + 3!) + 1$
$86 = 73 + 13$
$87 = \lceil{\sqrt{\sqrt{13^7}}}\rceil - 3 = \lceil{89.00222..}\rceil - 3$
$88 = 3^{3+1} + 7$
$89 = 13 \times 7 - \lceil\sqrt{3}\rceil$
$90 = 13 \times 7 - \lfloor\sqrt{3}\rfloor$
$91 = (3!+3!+1) \times 7$
$92 = 13 \times 7 + \lfloor\sqrt{3}\lfloor$
$93 = 13 \times 7 + \lceil\sqrt{3}\lceil$
$94 = 13 \times 7 + 3$
$95 = $
$96 = 17 \times 3! - 3!$
$97 = 73 + (3+1)!$
$98 = 71 + 3^3$
$99 = (3! - 1)! - 3\times 7$
$100 = 31 \times 3 + 7$

Bonus:

If we allow log functions we can generate every number like this:
$x = \log_{\frac{1}{\lfloor\sqrt7\rfloor}}\left({\log_3\underbrace{\sqrt{\sqrt{\dots\sqrt{3\,}\,}\,}}_\text{x square roots}}\right)$
This is equivalent to
$x = \log_{\frac12}\left({\log_3{3^{\frac{1}{2^x}}}}\right)$
Going further:
$x = \log_{\frac12}\left({\frac{1}{2^x}}\right)$

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  • $\begingroup$ I think for 78 and 79, you meant $3!\times (7+3!)\times 1$ and $3!\times (7+3!)+1$, right? $\endgroup$ – H_D Jan 29 at 14:18
  • $\begingroup$ You're welcome. It's kind of my responsibility to do so I guess? $\endgroup$ – H_D Jan 29 at 14:21
  • $\begingroup$ Very close! You can do it!! $\endgroup$ – H_D Jan 29 at 15:42
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    $\begingroup$ Here's a silly one for you: $95: 1+ \lfloor \sqrt{ \sqrt{ \sqrt{ \lfloor \sqrt{ 337 } \rfloor ! } } } \rfloor$ $\endgroup$ – Bass Jan 29 at 18:09
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Trying my best to arrange the numbers in order:

1-10:

1= $1*-(3+3-7)$
2= $1-(3+3-7)$
3= $13-3-7$
4= $1*floor(\sqrt{3})*-(3-7)$
5= $1^3*3+floor(\sqrt{7})$
6= $13*floor(\sqrt{3})-7$
7= $13+floor(\sqrt{3})-7$
8= $13-3-floor(7)$
9= $1*3^{\sqrt{(-3+7)}}$
10= $1+3^{\sqrt{(-3+7)}}$

11-20:

11= $13-floor(\sqrt{3})*floor(\sqrt{7})$
12= $13+floor(\sqrt{3})-floor(\sqrt{7})$
13= $13+ceil(\sqrt{3})-floor(\sqrt{7})$
14= $(1+3)!-3-7$
15= $13+floor(\sqrt{3})*floor(\sqrt{7})$
16= $(1+3)^{\sqrt{(-3+7)}}$
17= $13-3+7$
18= $1*3*3*floor(\sqrt{7})$
19= $13+3*floor(\sqrt{7})$
20= $(-1+3)*(3+7)$

21-30:

21= $1-floor(\sqrt{3})+3*7$
22= $1*floor(\sqrt{3})+3*7$
23= $1+floor(\sqrt{3})+3*7$
24= $1*3+3*7$
25= $(-1+3+3)^{floor(\sqrt{7})}$
26= $13*floor(\sqrt{3})*floor(\sqrt{7})$
27= $1*3*3*ceil(\sqrt{7})$
28= $-1+3^3+floor(\sqrt{7})$
29= $-1+3*(3+7)$
30= $1*3*(3+7)$

31-40:

31= $1+3*(3+7)$
32= $1*3^{3+floor(\sqrt{7})}$
33= $1+3^{3+floor(\sqrt{7})}$
34= $1*(-3)+37$
35= $1-3+37$
36= $-1*floor(\sqrt{3})+37$
37= $1*floor(\sqrt{3})+37$
38= $1+floor(\sqrt{3})+37$
39= $-1+3+37$
40= $1*3+37$

41-50:

41= $1+3+37$
42= $1*(3+3)*7$
43= $1+(3+3)*7$
44= $1+3!+37$
45= $(1+3)!+3*7$
46= $13*3+7$
47= $-1+3!+3!*7$
48= $(1+3)!*floor(\sqrt{3})*floor(\sqrt{7})$
49= $(1+3+3)*7$
50= $(-1+3!)*(3+7)$

51-60:

51= $1*floor(\sqrt{3!!})+floor(\sqrt{\sqrt{3!!}})^{floor(\sqrt{7})}$
52= $1+floor(\sqrt{3!!})+floor(\sqrt{\sqrt{3!!}})^{floor(\sqrt{7})}$
53= $1+ceil(\sqrt{3!!})+floor(\sqrt{\sqrt{3!!}})^{floor(\sqrt{7})}$
54= $1*3*3*floor(\sqrt{7})!$
55= $(-1+3!)(floor(\sqrt{\sqrt{3!!}}+ceil(\sqrt{7})!)$
56= $(-1+3!+3)*7$
57= $(-1+floor(\sqrt{3!!}))*ceil(\sqrt{3})+7$
58= $-(1+3)*3+floor(\sqrt{7!})$
59= $-13+ceil(\sqrt{3})+floor(\sqrt{7})$
60= $-13+3+floor(\sqrt{7!})$

61-70:

61= $-1*3*3+floor(\sqrt{7!})$
62= $-(1+3)*ceil(\sqrt{3})+floor(\sqrt{7!})$
63= $1*3*3*7$
64= $1+3*3*7$
65= $-1*3-ceil(\sqrt{3})+floor(\sqrt{7!})$
66= $-1-3-ceil(\sqrt{3})+floor(\sqrt{7!})$
67= $-1*3-floor(\sqrt{3})+floor(\sqrt{7!})$
68= $-1-floor(\sqrt{3})*floor(\sqrt{3})+floor(\sqrt{7!})$
69= $-1*3/3+floor(\sqrt{7!})$
70= $1*3/3*floor(\sqrt{7!})$

71-80:

71= $1*3/3+floor(\sqrt{7!})$
72= $-1+3*floor(\sqrt{3})+floor(\sqrt{7!})$
73= $1*3*floor(\sqrt{3})+floor(\sqrt{7!})$
74= $1+3*floor(\sqrt{3})+floor(\sqrt{7!})$
75= $-1+3*ceil(\sqrt{3})+floor(\sqrt{7!})$
76= $1*3*ceil(\sqrt{3})+floor(\sqrt{7!})$
77= $1+3*ceil(\sqrt{3})+floor(\sqrt{7!})$
78= $-1+3*3+floor(\sqrt{7!})$
79= $1*3*3+floor(\sqrt{7!})$
80= $1+3*3+floor(\sqrt{7!})$

81-90:

81= $-1+3!+3!+floor(\sqrt{7!})$
82= $1*3!+3!+floor(\sqrt{7!})$
83= $1+3!+3!+floor(\sqrt{7!})$
84= $-1+3^ceil(\sqrt{3})+floor(\sqrt{7!})$
85= $1*3^ceil(\sqrt{3})+floor(\sqrt{7!})$
86= $1+3^ceil(\sqrt{3})+floor(\sqrt{7!})$

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Extending Omega Krypton's answer with 8 to 15, keeping 1,3,3,7 in order. I use $\circ$ to display concatenation:

$8 = 1+3-3+7$
$9 = \lceil 1* \sqrt[3]{3}\rceil+7$
$10 = 1 \circ \lfloor 3/3 \circ 7 \rfloor$
$11 = 1\circ 3- \lceil \sqrt[3]{7} \rceil$
$12 = 1\circ 3- \lceil 3/7 \rceil$
$13 = 1\circ 3+ \lfloor 3/7 \rfloor$
$14 = 1\circ 3+ \lceil 3/7 \rceil$
$15 = 1\circ 3+ \lceil \sqrt[3]{7} \rceil$

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This puzzle is a bit too broad, so I only have three example numbers below. I'm sure you can figure out what the rest might look like.

$36 = \lfloor{\sqrt{1337}}\rfloor$

$11 = \lfloor{\sqrt{\lfloor{\sqrt{\lfloor{\sqrt{\lfloor{\sqrt{(\lfloor{\sqrt{\lfloor{\sqrt{\lfloor{\sqrt{\lfloor{\sqrt{\lfloor{\sqrt{(\lfloor{\sqrt{1337}}\rfloor)!}}\rfloor}}\rfloor}}\rfloor}}\rfloor}}\rfloor)!}}\rfloor}}\rfloor}}\rfloor}}\rfloor$
$79 = \lfloor{\sqrt{\lfloor{\sqrt{(\lfloor{\sqrt{\lfloor{\sqrt{\lfloor{\sqrt{\lfloor{\sqrt{(\lfloor{\sqrt{\lfloor{\sqrt{\lfloor{\sqrt{\lfloor{\sqrt{\lfloor{\sqrt{(\lfloor{\sqrt{1337}}\rfloor)!}}\rfloor}}\rfloor}}\rfloor}}\rfloor}}\rfloor)!}}\rfloor}}\rfloor}}\rfloor}}\rfloor)!}}\rfloor}}\rfloor$

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  • $\begingroup$ Surely I didn't think of this possibility. Well, fascinating. $\endgroup$ – H_D Jan 29 at 14:22
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Work in progress, I'm trying to avoid using ceil, floor and root, so far so good but I don't think it's possible for some big numbers. I could have made some mistakes though.

1-10:

$1=(1+3+3)/7$
$2=1+7-3-3$
$3=3*3-7+1$
$4=(31-3)/7$
$5=3!+3!-7/1$
$6=37-31$
$7=7+(3-3)*1$
$8=17-3*3$
$9=7+1+3/3$
$10=3!*3-7-1$

11-20:

$11=17-3-3$
$12=3!*3-7+1$
$13=7+3+3/1$
$14=7+3+3+1$
$15=7*3-3!/1$
$16=33-17$
$17=17-3+3$
$18=3*7-3/1$
$19=3*7-3+1$
$20=3!+3!+7+1$

21-30:

$21=31-7-3$
$22=17+3!-3$
$23=17+3+3$
$24=37-13$
$25=3*7+3+1$
$26=3!*3+7+1$
$27=(7-3)!+3/1$
$28=(7-3)!+3+1$
$29=3!*3!-7/1$
$30=3!*3!-7+1$

31-40:

$31=7*(3+1)+3$
$32=13*3-7$
$33=37-3-1$
$34=37-3/1$
$35=37-3+1$
$36=(7-1)*(3+3)$
$37=13+(7-3)!$
$38=71-33$
$39=37+3-1$
$40=37+3/1$

41-50:

$41=37+3+1$
$42=73-31$
$43=7*(3+3)+1$
$44=7*3!+3-1$
$45=7*3!+3/1$
$46=13*3+7$
$47=7*3!+3!-1$
$48=(7+1)*(3+3)$
$49=7*(3+3+1)$
$50=(7+3)*3!-1$

51-60: (WIP)

$51=$
$52=$
$53=$
$54=17*3+3$
$55=$
$56=$
$57=$
$58=$
$59=$
$60=(13+7)*3$

61-70: (WIP)

$61=$
$62=7*3*3-1$
$63=7*3*3/1$
$64=7*3*3+1$
$65=$
$66=$
$67=$
$68=37+31$
$69=73-3-1$
$70=(73-3)/1$

71-80: (WIP)

$71=73-3+1$
$72=71+3/3$
$73=$
$74=$
$75=$
$76=$
$77=71+3+3$
$78=$
$79=$
$80=$

81-90: (WIP)

$81=$
$82=$
$83=$
$84=$
$85=7*13-3!$
$86=31*3-7$
$87=$
$88=7*13-3$
$89=$
$90=$

91-100 (WIP)

$91=$
$92=$
$93=$
$94=7*13+3$
$95=$
$96=$
$97=7*13+3!$
$98=$
$99=$
$100=31*3+7$

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