- #31
Rogerio
- 404
- 1
Ooops,
66 = (3+8) \cdot \sqrt{4} \cdot \sqrt{9}
66 = (3+8) \cdot \sqrt{4} \cdot \sqrt{9}
Constructing Integers with 3,4,8,9 up to 101
- Context: Undergrad
- Thread starter ceptimus
- Start date
Click For Summary
Discussion Overview
The thread discusses the challenge of constructing all integers from 1 to 101 using the digits 3, 4, 8, and 9, employing various mathematical operations such as addition, subtraction, multiplication, division, and exponentiation. Participants share their solutions and methods, while also addressing the use of recurring decimals and the validity of certain approaches.
Discussion Character
- Exploratory
- Mathematical reasoning
- Debate/contested
Main Points Raised
- Participants propose various equations to represent integers, such as 1 = (4 - 3) / (9 - 8) and 2 = 4 + 9 - 3 - 8.
- Some participants express concerns about the validity of using certain operations, like the modulo function, which one participant describes as "cheating."
- There are multiple representations for the same integer, with different participants suggesting various forms, such as 19 = (7 × 3) - (8 / 4) and 19 = 34 - (8 + 7).
- Discussions about notation for recurring decimals arise, with suggestions for using symbols like .(3) or .\overline{3} to denote recurring values.
- Participants correct each other’s work and suggest alternative methods for constructing integers, indicating a collaborative effort to refine solutions.
- Some participants express frustration over specific integers that remain unsolved, such as 22 and 30, and seek assistance from others.
- There are playful exchanges about the use of advanced mathematical concepts, like square roots and factorials, with some participants humorously acknowledging their "cheating" in using these methods.
Areas of Agreement / Disagreement
The discussion features a mix of agreement on certain solutions and disagreement on the validity of specific methods. Participants do not reach a consensus on the best approaches or the use of certain mathematical operations, indicating that multiple competing views remain.
Contextual Notes
Some solutions involve unresolved mathematical steps or assumptions about the operations allowed, particularly regarding the use of recurring decimals and advanced functions. The scope of the challenge is limited to the digits provided and the specified operations.
Who May Find This Useful
Mathematics enthusiasts, educators, and students interested in number theory or combinatorial challenges may find this discussion engaging and informative.
- #32
The Bob
- 1,126
- 0
I was sure that ceptimus said no square root functions:
The Bob (2004 ©)
ceptimus said:
The Bob (2004 ©)
- #33
Rogerio
- 404
- 1
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 !
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 !
Last edited:
- #34
ceptimus
- 299
- 2
Rogerio said:
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.
Last edited:
- #35
Rogerio
- 404
- 1
ceptimus said:Nice.
Edit: I've not done a proper search, but I think we still need 58 and 83 too.
A fix:Rogerio said:
83 = 84 - .\overline{9} ^{ 3}
Yes! 67 is the last one!
I left it to Bob...
Last edited:
- #36
The Bob
- 1,126
- 0
This is all I can get: 67 = (8\times9) - \sqrt{4} -3Rogerio said:I left it to Bob...
I am rubbish at this. Sorry. I will think some more.
The Bob (2004 ©)
- #37
ceptimus
- 299
- 2
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.
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.
- #38
dextercioby
Science Advisor
- 13,404
- 4,184
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.
,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.
- #39
Rogerio
- 404
- 1
ceptimus said:
One of them is (probably) :
67 = \frac{9}{.3 \times .4} - 8
... and the other one?
Last edited:
- #40
Rogerio
- 404
- 1
...what about 102 ? Is there a way, ceptimus ?
- #41
Rogerio
- 404
- 1
Rogerio said:
Well, I think it could be
65 = ( 8 - .\overline{3} - .\overline{4} ) \times 9
OOOPS ! it should be 67 !
but... and 102 ?
Last edited:
- #42
ceptimus
- 299
- 2
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.
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.
- #43
Rogerio
- 404
- 1
ceptimus said:
Then add 121 to the collection :-)
121 = \frac{8}{.4 - .\overline{3}} + .\overline{9}
- #44
ceptimus
- 299
- 2
Here's the difficult 67 as promised:
67 = \frac{.3 + .\overline{4}}{.9 - .\overline{8}}
Happy New Year.
67 = \frac{.3 + .\overline{4}}{.9 - .\overline{8}}
Happy New Year.
Similar threads
- · Replies 4 ·
High School
Two interesting number theory tricks
- · Replies 3 ·
- · Replies 3 ·
- · Replies 4 ·
- · Replies 1 ·
- · Replies 2 ·
- · Replies 5 ·
- · Replies 6 ·
- · Replies 2 ·