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!