Constructing Integers with 3,4,8,9 up to 101

[ceptimus]

Using each time all of the four digits: 3, 4, 8 and 9 construct all the integers up to and including 101. You may only use addition, subtraction, multiplication, division, the decimal point, raising to a power, and (if there is no other way) a recurring decimal. Decimal fractions without a leading integer are allowed. You can also run digits together to make numbers like 9 = 48 - 39

Factorials are not allowed, nor is the square root operation, and you can't use constructions like (3 + 4)8 to mean 78.

OK, I'll start you off:

1 = \frac{4 - 3}{9 - 8}

2 = 4 + 9 - 3 - 8

3 = \frac{4}{9.3(recurring) - 8}

I know there are easier ways to make 3, but I wanted to show what I meant by recurring.

Have fun.

 

[Galileo]

4=9-3-\frac{8}{4}

5=\frac{8}{4}+\frac{9}{3}
6=3+4+8-9

7=(4+3)(9-8)

8=3+4+9-8

9=9(8-4-3)

10=9+8-3-4

Go me!

 

[Galileo]

11 = (3)(4)+8-9

12 = (3)(4)(9-8)

13 = (3)(4)+9-8

14 = 9+3+\frac{8}{4}

15 = \frac{(8)(3)}{4}+9

16 = 9+8+3-4

17 = (9+8)(4-3)

18 = 9+8+4-3

19 = (3)(8)+4-9

20 = \frac{9}{3}(4)+8

 

[Gokul43201]

11=49-38


12=3*4*(9-8)


13=(3*4)+9-8


14=9+3+\frac{8}{4}

15=8+4+\frac{9}{3}


16=9+8-4+3


17=34-9-8


18=8+9+4-3

19=4!-8+\frac{9}{3} (cheater !)

20=(\frac{4}{3}*9)+8


Someone please fix 19 for me.

 

[rachmaninoff]

19=(7\times3)-\frac{8}{4}

19=34-(8+7)

19=3^{7-4}-8

------------------------------------------

21=(8\times3)+4-7

22=(8\times4)-(7+3)

23=\frac{8}{4}+(7\times3)

24=7^3\bmod{4}*8

25=(7\times3)+8-4

26=\frac{38}{4}+7

27=38-(4+7)

28=43-(8+7)

29=(8+4)\times3-7

30=(7\times4)+8\bmod{3}

 

[ceptimus]

Nice work guys. But that mod function is cheating.

 

[RandallB]

11 =49-38 =3*8 - 9 - 4 =3*4 +8 -9
12= (9-3)(8/4)
13= 3*4 +9 -8
14= 9+3+8/4
15= (9)(8/4)-3 = (3/4)*8+9 =9/3+4+8= 3*9 -4 -8
16= 3+8+9-4
17= 34-9-8=(4^3)/8 +9
18= +4 +8 +9 -3 =(9-3)*4-8
19= 3*8 +4 -9
20 = (4/3)*9 +8

 

[RandallB]

19=(7\times3)-\frac{8}{4}

19=34-(8+7)

30=(7\times4)+8\bmod{3}

He want's 9's not 7's

 

[dextercioby]

41=9\cdot 4+8-3
42=43-9+8
44=43-8+9=8\cdot 4+9+3=(8+\frac{9}{3})\cdot 4
46=89-43
47=9\cdot 4+8+3
51=34+8+9
54=\frac{9\cdot 8\cdot 3}{4}=\frac{9\cdot 8}{.(4)\cdot 3}
55=89-34=98-43
56=4^{\frac{9}{3}}-8
59=8\cdot 4+9\cdot 3
60=43+8+9=8\cdot 9-3\cdot 4=8\cdot 3+9\cdot 4
65=8\cdot 9-4-3=98-34
71=8\cdot 9-4+3=8.(3)\cdot 9-4
72=\frac{4^{3}}{8}\cdot 9=4^{\frac{9}{3}}+8
73=8\cdot 9-3+4
76=(9\cdot 3-8)\cdot 4
79=8\cdot 9+3+4
81=9\cdot 3^{\frac{8}{4}}
82=89-3-4
84=8\cdot 9+3\dot 4
87=(\frac{3}{.(4)}\cdot 8)-9
88=89-4+3
90=89-3+4
96=89+3+4=\frac{9\cdot 8\cdot 4}{3}=.(4)\cdot 3\cdot 9\cdot 8=.(3)\cdot 4\cdot 8\cdot 9
97=98-4+3
99=98+4-3
100=\frac{3}{.(4)}\cdot 9-8

Daniel.

Fill in the gaps

 

[ceptimus]

19=\frac{8}{.3}+4-9

30=\frac{8}{.4}+9+3

I think we need to put in the (recurring) or some symbol that means that. I think Daniel's solution for 30, as written, is actually a solution for 32.

No doubt one of you LaTeX gurus will know the correct formatting for a recurring decimal.

 

[dextercioby]

I think we need to put in the (recurring) or some symbol that means that. I think Daniel's solution for 30, as written, is actually a solution for 32.

No doubt one of you LaTeX gurus will know the correct formatting for a recurring decimal.

I feel like a jackass. I erased the message by mistake.
19=\frac{8}{.(3)}+4-9
20=\frac{9\cdot 8}{3}-4=8\cdot 4-9-3
21=\frac{8}{.(4)}+\frac{9}{3}
23=9\cdot 3-8+4=\frac{8}{.4}+\frac{9}{3}
24=\frac{8}{.(4)}+9-3
25=9\cdot 3-\frac{8}{4}
26=\frac{8}{.4}+9-3=8\cdot 4-9+3
28=\frac{9\cdot 8}{3}+4=\frac{9}{.3}-\frac{8}{4}
29=8\cdot 3+9-4
30=\frac{8}{.(4)}+9+3
31=9\cdot 3+8-4
32=\frac{8}{.4}+9+3=\frac{9}{.3}+\frac{8}{4}

33=34-9+8
35=34-8+9=8\cdot 4+\frac{9}{3}
37=8\cdot 3+9+4
38=8\cdot 4+9-3
39=9\cdot 3+8+4

Daniel.
Fill in the gaps.
PS.I may have not transcripted everything i had before. I'll be a guru next year.Guru of physics...

 

[rachmaninoff]

He want's 9's not 7's

oops?

Btw, is this notation any good?

19=\frac{8}{.\overline{3}} + 4 - 9

 

[dextercioby]

oops?

Btw, is this notation any good?

19=\frac{8}{.\overline{3}} + 4 - 9

In my fifth grade (11 years ago ) i learned that:
4,33333333333...=4.(3)=4\frac{3}{9}=4\frac{1}{3}=\frac{13}{3}

So,for me,it's obvious the notation.And besides why write ".\overline{3}" when u can easily put two brackets:".(3)"...?

Daniel.

 

[dextercioby]

Some more

64=3^{4}-9-8
80=3^{4}-9+8
82=3^{4}-8+9
98=3^{4}+9+8

72=\frac{3^{4}}{9}\cdot 8
69=\frac{3}{.4}\cdot 8+9
51=\frac{3}{.4}\cdot 8-9
81=9^{4-\sqrt[3]{8}} CHEATING!

Daniel.


PS.Ceptimus,it's not fair!Let us use at least \sqrt[3]{8} or \sqrt{9};\sqrt{4}.Please...

 

[Galileo]

Just for clarity. I've attempted to fill the gaps for 21-30.
And I don't really like using the recurring decimal or decimal point. 22 and 30 are still missing non-decimal form.

21=3(9-\frac{8}{4})

22=..?

23=(9)(3)+4-8

24=3+4+8+9

25=(3)(9)-\frac{8}{4}

26=(8)(4)+3-9

27=3^{4-(9-8))

28=\frac{9}{3}8+4

29=(3)(4)+8+9

30=..??

 

[dextercioby]

Since that 22 is really eataing me alive,i decided to take the sword and cut the Gordian knot.
87=48+39=49+38
22=(\sqrt[3]{8}\cdot 9)+4=8\cdot 3-[\frac{9}{4}]=[\frac{9}{.(8)}]+4\cdot 3=(9+\sqrt[3]{8})\cdot\sqrt{4}=...


Daniel.

 

[Gokul43201]

Ha ha...using the greatest integer function is one quick way to cut a Gordian Knot all right !

 

[Rogerio]

22 = 9\cdot 3 - 4/.8

30 = 4\cdot 8-3+.\overline{9}

87 = 3 \cdot 4 \cdot 8 - 9

81 = \frac{(84 - 3)} {.\overline{9}}

101 = 89+ \frac{3} {.4}

 

[Rogerio]

43 = \frac{43} {9-8}

45 = 93 - 48

48 = (3+9) \cdot (8-4)

49 = (3+4) \cdot (8-.\overline{9})

50 = 48+3-.\overline{9}

52 = 48+3+.\overline{9}

57 = (3+4) \cdot 8 + .\overline{9}

63 = (3+8-4) \cdot 9

70 = \frac{3+4} {.9-.8}

77 = 89 - \frac{4}{.\overline{3}}

78 = 39 \cdot \frac{8}{4}

86 = 98 - 3 \cdot 4

89 = \frac{89}{4-3}

91 = 98 - 3 - 4

 

[Rogerio]

Someone required to play with 53 58 61 62 66 67 68 74 75 83 85 92 93 94 :-)

 

[ceptimus]

101 = 89+ \frac{3} {.4}

I've not checked all of them, but this one looks wrong:

3 / .4 = 7.5
3 / .444... = 6.75

I suppose you meant 4 / .(3)

However 101 = 89 + 3 \times 4

 

[Rogerio]

I've not checked all of them, but this one looks wrong:
101 = 89+ \frac{3} {.4}

I suppose you meant 4 / .(3)

No, I meant 89 + 3 \cdot 4

The frac was a cut&paste error :-)

 

[Rogerio]

A little bit more:

53 = \frac{3 \cdot 8}{.\overline{4}} - .\overline{9}

61 = \frac{3 \cdot 8}{.4} + .\overline{9}

92 = (3 \cdot 8 - .\overline{9}) \cdot 4

93 = (4 \cdot 8 - .\overline{9}) \cdot 3

 

[NateTG]

81=9^{4-\sqrt[3]{8}} CHEATING!

Daniel.


PS.Ceptimus,it's not fair!Let us use at least \sqrt[3]{8} or \sqrt{9};\sqrt{4}.Please...

Isn't this just an excuse to get fancy?
81=\left(\frac{9}{3}\right)^{8-4}

 

[TD]

((4+9)*8)-3 = 101
((9+4)*8)-3 = 101

 

[Rogerio]

A little step:

74 = 38 + 4 \cdot 9

75 = 93 - \frac{8}{.\overline{4}}

94 = \frac{38}{.4} - .\overline{9}


Resting 58 62 66 67 68 83 85 ...:-)

 

[dextercioby]

Yeah,Nate,u're right,but it makes life easier...
67=8^{[\frac{9}{4}]}+3
66=8^{[\sqrt{3}]}+[\frac{9}{4}]
62=8^{[\sqrt{3}]}-[\frac{9}{4}]
58=8^{\sqrt{4}}-3-\sqrt{9}
85=9^{[\sqrt{3}]}+8-4
83=9^{\sqrt{4}}+\sqrt[3]{8}

Sometimes cheating makes u winner.

Daniel.

PS.You're right about 81,though:
81=3^{4(9-8)}

EDIT:A mistake on the first page:
100=3\cdot 4\cdot 9-8

 

[Rogerio]

Resting 58 62 66 67 68 83 85 ...:-)

62 = \frac{9}{.3} + 4 \cdot 8

Only 58 66 67 68 83 85

 

[Rogerio]

Only 58 66 67 68 83 85

who did say that?!


58 = \frac{9 + 8.4}{.3}

Just 66 67 68 83 85

 

[Rogerio]

I give up. I need "power" !


66 = 11 \cdot \sqrt{4} \cdot \sqrt{9}

67 = 8^{\sqrt{4}} + \frac{3}{.\overline{9}}

68 = 34 \cdot \sqrt [\sqrt{9}] {8}

85 = (8+9) \cdot (3+\sqrt{4})

Done!

 

[Rogerio]

Ooops,
66 = (3+8) \cdot \sqrt{4} \cdot \sqrt{9}

 

[The Bob]

I was sure that ceptimus said no square root functions:

Factorials are not allowed, nor is the square root operation, and you can't use constructions like (3 + 4)8 to mean 78.

The Bob (2004 ©)

 

[Rogerio]

Ok Bob...then

66 = 8 \cdot ( 9 - \frac{3}{4} )

68 = 8 ^ { ( 3 - .\overline{9} )} + 4

85 = 9 ^ { ( 8 ^ { .\overline{3} } )} + 4


...and I left the "67" for you !

 

[ceptimus]

85 = 9 ^ { ( 8 ^ { .\overline{3} } )} + 4

Nice. Is 67 the last one left then?

Edit: I've not done a proper search, but I think we still need 58 and 83 too.

 

[Rogerio]

Nice. Is 67 the last one left then?

Edit: I've not done a proper search, but I think we still need 58 and 83 too.

58 = \frac{9 + 8.4}{.3}

A fix:

83 = 84 - .\overline{9} ^{ 3}

Yes! 67 is the last one!

I left it to Bob...

 

[The Bob]

I left it to Bob...

This is all I can get: 67 = (8\times9) - \sqrt{4} -3

I am rubbish at this. Sorry. I will think some more.

The Bob (2004 ©)

 

[ceptimus]

There are two solutions (I know of) for 67.

One is very hard, and uses two recurring decimals. The other uses the decimal point but no recurring decimals.

I'll post them on New Year's Eve if no one gets them before and no one tells me not to.

 

[dextercioby]

67=\frac{8}{0.(3)|}-4.(9)
,where "|" is a (perfectly reflectant) mirror,and the multiplicative dot '\cdot' is understood.

Daniel.

PS.Ceptimus,u didn't say anything about "mirrors",right?It has 2 recurring decimals and adding the image,three in total.

 

[Rogerio]

There are two solutions (I know of) for 67.

One of them is (probably) :

67 = \frac{9}{.3 \times .4} - 8

... and the other one?

 

[Rogerio]

...what about 102 ? Is there a way, ceptimus ?

 

[Rogerio]

... and the other one?

Well, I think it could be

65 = ( 8 - .\overline{3} - .\overline{4} ) \times 9

OOOPS ! it should be 67 !

but... and 102 ?

 

[ceptimus]

I don't have answers for 102 or 109. All the others up to 120 I have. I've not searched beyond 120.

Well done on the 67 solution. I won't reveal the other (harder) 67 answer until New Year's Eve, just in case someone is working on it.

 

[Rogerio]

I've not searched beyond 120...

Then add 121 to the collection :-)

121 = \frac{8}{.4 - .\overline{3}} + .\overline{9}

 

[ceptimus]

Here's the difficult 67 as promised:

67 = \frac{.3 + .\overline{4}}{.9 - .\overline{8}}

Happy New Year.