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[Drozmight]

Anyone ever try and figure out that Pedals Around The Rose game?

Took me 20 mintues to figure it out...

Then I was ... :| Lame.

Spoiler

The way they put THE in the name really throws you off... I thought the five die represented a rose.

 

[mashoutposse]

7 is pretty much impossible.

7+2 = 9, 7-2 = 5, 7*2 = 14, 7/2 = 3.5.

Since none of those numbers can be generated with three 2s, 7 can't be generated with four -- unless you "cheat;" 2/(.2*2) = 5.

 

[RonaldoSan]

1 = ((2/2)+2)-2
2 = (2/2)+(2/2)
3 = (2+2+2)/2
4 = ((2*2)*2)/2
5 = (2+2)+(2/2)
6 = ((2/2)+2)*2
7 = ?
8 = 2+2+2+2
9 = (22/2)-2
10 = ((2*2)*2)+2

 

[GaimeGuy]

Hmm...

Maybe its like programming?

(2++) + (2++) + (2/2) = 7 ??

HOORAY FOR GEEKS

 

[nitewulf]

i believe the comma is the british standard for the decimal point, and as such ".2" is valid. aside from that, there's no way to get 7, im pretty certain.

 

[Meatpuppet]

Sorry, but it's not.

 

[nitewulf]

well, we learned that way when i was a kid. must be a pretty old system that was kept around the subcontinent till fairly recently.

 

[Meatpuppet]

That system is used in a number of countries in continental Europe. France is one. However, the English speaking ones don't.

We also drive on the wrong side of the road.

 

[Mister Zimbu]

2() - /() - 2 - (2/2)

I win!

 

[Poody]

2() - /() - 2 - (2/2)

I win!

huh?

 

[Justin Bailey]

huh?

Just nod your head slowly and pat him on the back.

 

[Hitokage]

20 - 10 - 2 - (2/2)

is what he means.

 

[LizardKing]

despite the people talking about antiquated systems and using a comma, he never said we could use a comma

 

[LordPhlegm]

 

[GaimeGuy]

There is no solution for 7.

 

[LordPhlegm]

There is no solution for 7.

I came to the same conclusion that all of you had about 7 being impossible. But, in a physical sense, I'd imagine my answer is legit.

 

[Raistlin]

Wow ... there are A LOT of ways to do it with five 2's, but four appears impossible ...





Let's see how many ways we can do it with five 2's??!!

 

[shattyboombatty]

Well someone had to do it... Here's the output of a C program I trew together. It's every combination of 4 2's and 4 operators. You'll see no 7's as an answer. I left it as float answers so as to show fractional answers like .5 if they came up. I'm assuming that since '(' ')' just determine order of operations, all the operations have the same presedence (in this program) and I've got all the combinations that I can leave out the parenthesation of all the terms. (let me know if that's wrong) Plus I didn't want to program the grouping logic. =)

Also, the combining (22 222) logic may have bugs but I think these are correct =)

1) 2 + 2 + 2 + 2 = 8.0
2) 2 + 2 + 2 - 2 = 4.0
3) 2 + 2 + 2 * 2 = 12.0
4) 2 + 2 + 2 / 2 = 3.0
5) 2 + 2 + 22 = 26.0
6) 2 + 2 - 2 + 2 = 4.0
7) 2 + 2 - 2 - 2 = 0.0
8) 2 + 2 - 2 * 2 = 4.0
9) 2 + 2 - 2 / 2 = 1.0
10) 2 + 2 - 22 = -18.0
11) 2 + 2 * 2 + 2 = 10.0
12) 2 + 2 * 2 - 2 = 6.0
13) 2 + 2 * 2 * 2 = 16.0
14) 2 + 2 * 2 / 2 = 4.0
15) 2 + 2 * 22 = 88.0
16) 2 + 2 / 2 + 2 = 4.0
17) 2 + 2 / 2 - 2 = 0.0
18) 2 + 2 / 2 * 2 = 4.0
19) 2 + 2 / 2 / 2 = 1.0
20) 2 + 2 / 22 = 0.1
21) 2 + 22 + 2 = 26.0
22) 2 + 22 - 2 = 22.0
23) 2 + 22 * 2 = 48.0
24) 2 + 22 / 2 = 12.0
25) 2 + 222 = 224.0
26) 2 - 2 + 2 + 2 = 4.0
27) 2 - 2 + 2 - 2 = 0.0
28) 2 - 2 + 2 * 2 = 4.0
29) 2 - 2 + 2 / 2 = 1.0
30) 2 - 2 + 22 = 22.0
31) 2 - 2 - 2 + 2 = 0.0
32) 2 - 2 - 2 - 2 = -4.0
33) 2 - 2 - 2 * 2 = -4.0
34) 2 - 2 - 2 / 2 = -1.0
35) 2 - 2 - 22 = -22.0
36) 2 - 2 * 2 + 2 = 2.0
37) 2 - 2 * 2 - 2 = -2.0
38) 2 - 2 * 2 * 2 = 0.0
39) 2 - 2 * 2 / 2 = 0.0
40) 2 - 2 * 22 = 0.0
41) 2 - 2 / 2 + 2 = 2.0
42) 2 - 2 / 2 - 2 = -2.0
43) 2 - 2 / 2 * 2 = 0.0
44) 2 - 2 / 2 / 2 = 0.0
45) 2 - 2 / 22 = 0.0
46) 2 - 22 + 2 = -18.0
47) 2 - 22 - 2 = -22.0
48) 2 - 22 * 2 = -40.0
49) 2 - 22 / 2 = -10.0
50) 2 - 222 = -220.0
51) 2 * 2 + 2 + 2 = 8.0
52) 2 * 2 + 2 - 2 = 4.0
53) 2 * 2 + 2 * 2 = 12.0
54) 2 * 2 + 2 / 2 = 3.0
55) 2 * 2 + 22 = 26.0
56) 2 * 2 - 2 + 2 = 4.0
57) 2 * 2 - 2 - 2 = 0.0
58) 2 * 2 - 2 * 2 = 4.0
59) 2 * 2 - 2 / 2 = 1.0
60) 2 * 2 - 22 = -18.0
61) 2 * 2 * 2 + 2 = 10.0
62) 2 * 2 * 2 - 2 = 6.0
63) 2 * 2 * 2 * 2 = 16.0
64) 2 * 2 * 2 / 2 = 4.0
65) 2 * 2 * 22 = 88.0
66) 2 * 2 / 2 + 2 = 4.0
67) 2 * 2 / 2 - 2 = 0.0
68) 2 * 2 / 2 * 2 = 4.0
69) 2 * 2 / 2 / 2 = 1.0
70) 2 * 2 / 22 = 0.18
71) 2 * 22 + 2 = 46.0
72) 2 * 22 - 2 = 42.0
73) 2 * 22 * 2 = 88.0
74) 2 * 22 / 2 = 22.0
75) 2 * 222 = 444.0
76) 2 / 2 + 2 + 2 = 5.0
77) 2 / 2 + 2 - 2 = 1.0
78) 2 / 2 + 2 * 2 = 6.0
79) 2 / 2 + 2 / 2 = 1.5
80) 2 / 2 + 22 = 23.0
81) 2 / 2 - 2 + 2 = 1.0
82) 2 / 2 - 2 - 2 = -3.0
83) 2 / 2 - 2 * 2 = -2.0
84) 2 / 2 - 2 / 2 = -0.5
85) 2 / 2 - 22 = -21.0
86) 2 / 2 * 2 + 2 = 4.0
87) 2 / 2 * 2 - 2 = 0.0
88) 2 / 2 * 2 * 2 = 4.0
89) 2 / 2 * 2 / 2 = 1.0
90) 2 / 2 * 22 = 22.0
91) 2 / 2 / 2 + 2 = 2.5
92) 2 / 2 / 2 - 2 = -1.5
93) 2 / 2 / 2 * 2 = 1.0
94) 2 / 2 / 2 / 2 = 0.3
95) 2 / 2 / 22 = 0.0
96) 2 / 22 + 2 = 2.0
97) 2 / 22 - 2 = -1.9
98) 2 / 22 * 2 = 0.1
99) 2 / 22 / 2 = 0.0
100) 2 / 222 = 0.0
101) 22 + 2 + 2 = 26.0
102) 22 + 2 - 2 = 22.0
103) 22 + 2 * 2 = 48.0
104) 22 + 2 / 2 = 12.0
105) 22 + 22 = 44.0
106) 22 - 2 + 2 = 22.0
107) 22 - 2 - 2 = 18.0
108) 22 - 2 * 2 = 40.0
109) 22 - 2 / 2 = 10.0
110) 22 - 22 = 0.0
111) 22 * 2 + 2 = 46.0
112) 22 * 2 - 2 = 42.0
113) 22 * 2 * 2 = 88.0
114) 22 * 2 / 2 = 22.0
115) 22 * 22 = 484.0
116) 22 / 2 + 2 = 13.0
117) 22 / 2 - 2 = 9.0
118) 22 / 2 * 2 = 22.0
119) 22 / 2 / 2 = 5.5
120) 22 / 22 = 1.0
121) 222 + 2 = 224.0
122) 222 - 2 = 220.0
123) 222 * 2 = 444.0
124) 222 / 2 = 111.0
125) 2222 = 2222.0
Press any key to continue

 

[DDayton]

There are two options...

1) The daughter didn't reveal all of the puzzle requirements.

2) The daughter made a mistake.

3) The daughter made up the challenge.

4) There is some incredibly evil twist we aren't getting... I was pondering using Roman numerals to do this, or dots on paper...

 

[shattyboombatty]

BTW I see some combinations I probably left out, but I'm only going to try so hard to do a little girls homework. =P

 

[Raistlin]

Well someone had to do it... Here's the output of a C program I trew together. It's every combination of 4 2's and 4 operators. You'll see no 7's as an answer. I left it as float answers so as to show fractional answers like .5 if they came up. I'm assuming that since '(' ')' just determine order of operations, all the operations have the same presedence (in this program) and I've got all the combinations that I can leave out the parenthesation of all the terms. (let me know if that's wrong) Plus I didn't want to program the grouping logic. =)

Also, the combining (22 222) logic may have bugs but I think these are correct =)

1) 2 + 2 + 2 + 2 = 8.0
2) 2 + 2 + 2 - 2 = 4.0
3) 2 + 2 + 2 * 2 = 12.0
4) 2 + 2 + 2 / 2 = 3.0
5) 2 + 2 + 22 = 26.0
6) 2 + 2 - 2 + 2 = 4.0
7) 2 + 2 - 2 - 2 = 0.0
8) 2 + 2 - 2 * 2 = 4.0
9) 2 + 2 - 2 / 2 = 1.0
10) 2 + 2 - 22 = -18.0
11) 2 + 2 * 2 + 2 = 10.0
12) 2 + 2 * 2 - 2 = 6.0
13) 2 + 2 * 2 * 2 = 16.0
14) 2 + 2 * 2 / 2 = 4.0
15) 2 + 2 * 22 = 88.0
16) 2 + 2 / 2 + 2 = 4.0
17) 2 + 2 / 2 - 2 = 0.0
18) 2 + 2 / 2 * 2 = 4.0
19) 2 + 2 / 2 / 2 = 1.0
20) 2 + 2 / 22 = 0.1
21) 2 + 22 + 2 = 26.0
22) 2 + 22 - 2 = 22.0
23) 2 + 22 * 2 = 48.0
24) 2 + 22 / 2 = 12.0
25) 2 + 222 = 224.0
26) 2 - 2 + 2 + 2 = 4.0
27) 2 - 2 + 2 - 2 = 0.0
28) 2 - 2 + 2 * 2 = 4.0
29) 2 - 2 + 2 / 2 = 1.0
30) 2 - 2 + 22 = 22.0
31) 2 - 2 - 2 + 2 = 0.0
32) 2 - 2 - 2 - 2 = -4.0
33) 2 - 2 - 2 * 2 = -4.0
34) 2 - 2 - 2 / 2 = -1.0
35) 2 - 2 - 22 = -22.0
36) 2 - 2 * 2 + 2 = 2.0
37) 2 - 2 * 2 - 2 = -2.0
38) 2 - 2 * 2 * 2 = 0.0
39) 2 - 2 * 2 / 2 = 0.0
40) 2 - 2 * 22 = 0.0
41) 2 - 2 / 2 + 2 = 2.0
42) 2 - 2 / 2 - 2 = -2.0
43) 2 - 2 / 2 * 2 = 0.0
44) 2 - 2 / 2 / 2 = 0.0
45) 2 - 2 / 22 = 0.0
46) 2 - 22 + 2 = -18.0
47) 2 - 22 - 2 = -22.0
48) 2 - 22 * 2 = -40.0
49) 2 - 22 / 2 = -10.0
50) 2 - 222 = -220.0
51) 2 * 2 + 2 + 2 = 8.0
52) 2 * 2 + 2 - 2 = 4.0
53) 2 * 2 + 2 * 2 = 12.0
54) 2 * 2 + 2 / 2 = 3.0
55) 2 * 2 + 22 = 26.0
56) 2 * 2 - 2 + 2 = 4.0
57) 2 * 2 - 2 - 2 = 0.0
58) 2 * 2 - 2 * 2 = 4.0
59) 2 * 2 - 2 / 2 = 1.0
60) 2 * 2 - 22 = -18.0
61) 2 * 2 * 2 + 2 = 10.0
62) 2 * 2 * 2 - 2 = 6.0
63) 2 * 2 * 2 * 2 = 16.0
64) 2 * 2 * 2 / 2 = 4.0
65) 2 * 2 * 22 = 88.0
66) 2 * 2 / 2 + 2 = 4.0
67) 2 * 2 / 2 - 2 = 0.0
68) 2 * 2 / 2 * 2 = 4.0
69) 2 * 2 / 2 / 2 = 1.0
70) 2 * 2 / 22 = 0.18
71) 2 * 22 + 2 = 46.0
72) 2 * 22 - 2 = 42.0
73) 2 * 22 * 2 = 88.0
74) 2 * 22 / 2 = 22.0
75) 2 * 222 = 444.0
76) 2 / 2 + 2 + 2 = 5.0
77) 2 / 2 + 2 - 2 = 1.0
78) 2 / 2 + 2 * 2 = 6.0
79) 2 / 2 + 2 / 2 = 1.5
80) 2 / 2 + 22 = 23.0
81) 2 / 2 - 2 + 2 = 1.0
82) 2 / 2 - 2 - 2 = -3.0
83) 2 / 2 - 2 * 2 = -2.0
84) 2 / 2 - 2 / 2 = -0.5
85) 2 / 2 - 22 = -21.0
86) 2 / 2 * 2 + 2 = 4.0
87) 2 / 2 * 2 - 2 = 0.0
88) 2 / 2 * 2 * 2 = 4.0
89) 2 / 2 * 2 / 2 = 1.0
90) 2 / 2 * 22 = 22.0
91) 2 / 2 / 2 + 2 = 2.5
92) 2 / 2 / 2 - 2 = -1.5
93) 2 / 2 / 2 * 2 = 1.0
94) 2 / 2 / 2 / 2 = 0.3
95) 2 / 2 / 22 = 0.0
96) 2 / 22 + 2 = 2.0
97) 2 / 22 - 2 = -1.9
98) 2 / 22 * 2 = 0.1
99) 2 / 22 / 2 = 0.0
100) 2 / 222 = 0.0
101) 22 + 2 + 2 = 26.0
102) 22 + 2 - 2 = 22.0
103) 22 + 2 * 2 = 48.0
104) 22 + 2 / 2 = 12.0
105) 22 + 22 = 44.0
106) 22 - 2 + 2 = 22.0
107) 22 - 2 - 2 = 18.0
108) 22 - 2 * 2 = 40.0
109) 22 - 2 / 2 = 10.0
110) 22 - 22 = 0.0
111) 22 * 2 + 2 = 46.0
112) 22 * 2 - 2 = 42.0
113) 22 * 2 * 2 = 88.0
114) 22 * 2 / 2 = 22.0
115) 22 * 22 = 484.0
116) 22 / 2 + 2 = 13.0
117) 22 / 2 - 2 = 9.0
118) 22 / 2 * 2 = 22.0
119) 22 / 2 / 2 = 5.5
120) 22 / 22 = 1.0
121) 222 + 2 = 224.0
122) 222 - 2 = 220.0
123) 222 * 2 = 444.0
124) 222 / 2 = 111.0
125) 2222 = 2222.0
Press any key to continue

Unfortunately, parenthesis are needed since the available operators do not have the same precedence. Also, the lack of parenthesis removes non-subtraction negation, which should be available.

One other thing - I don't there is anything against exponentiation (whether positive of negative), though I don't know if that will help with the missing 7 issue. It does allow for a number of alternative solutions for the other numbers however.

The algorithm is actually somewhat complicated when including the above possibilities.

 

[shattyboombatty]

Unfortunately, parenthesis are needed since the available operators do not have the same precedence. Also, the lack of parenthesis removes non-subtraction negation, which should be available.

One other thing - I don't there is anything against exponentiation (whether positive of negative), though I don't know if that will help with the missing 7 issue. It does allow for a number of alternative solutions for the other numbers however.

The algorithm is actually somewhat complicated when including the above possibilities.

Yeah, negation was what I forgot to consider. And if I'm just performing the operations from left to right then they do have the same presedency... although that's not normally the case.

I think exponentiation has an operator of essentially ^ which is implied by the exponents' position... but it'd be pretty sneaky to put it in there. =)

 

[DDayton]

I think exponentiation has an operator of essentially ^ which is implied by the exponents' position... but it'd be pretty sneaky to put it in there. =)

However, in standard written equations, there is no "operator" being used when writing exponential values. Heh.

 

[shattyboombatty]

However, in standard written equations, there is no "operator" being used when writing exponential values. Heh.

yeah, thus the "implied" and "sneaky"

 

[Aruarian Reflection]

There are two options...

1) The daughter didn't reveal all of the puzzle requirements.

2) The daughter made a mistake.

3) The daughter made up the challenge.

4) There is some incredibly evil twist we aren't getting... I was pondering using Roman numerals to do this, or dots on paper...

Notice that the OP hasn't come back to post. I'm guessing he's laughing at us at the moment. There is no daughter, and there is no homework.

 

[Raistlin]

I blame PI.



22 / 7 is an approximation of it

 

[LordPhlegm]

From before, any input on this answer at all?

Taking a sheet of paper and a strong black marker (or just using a transparency and a marker), writing three 2's on one side and one 2 on the other side can lead to this answer.

Furthermore, I think MashoutPosse explained a simple way of determing whether the given functions can result with a 7. No need for heavy thinking, programming, or anything complicated at all.

7 is pretty much impossible.

7+2 = 9, 7-2 = 5, 7*2 = 14, 7/2 = 3.5.

Since none of those numbers can be generated with three 2s, 7 can't be generated with four -- unless you "cheat;" 2/(.2*2) = 5.

If considering powers of 2, then consider the square root of 7 or the square root of 1 over 7, which really won't lead anywhere. Thus, four 2 units can't have an output of a seven with the given rules. Which leads me to believe, if there is an answer, it'll involve (excuse the cliche) thinking outside of the box. Anyway, if there is an answer and it's not mine, hope you guys find it.

 

[Stuggernaut]

Yeah I am starting to think that either ...

A) the answer uses some lame trick or play on words/numbers

B) My daughter has left something out of the explanation.

Either way...it's annoying!

 

[DDayton]

I think I'm happy with these two answers:

LordPhlegm:

Taking a sheet of paper and a strong black marker (or just using a transparency and a marker), writing three 2's on one side and one 2 on the other side can lead to this answer.


mashoutposse: 2/(.2*2) = 5

I prefer the reverse image solution -- sounds very Carrollian.

 

[hobart]

Check this out:

2^() + (2+2+2) = 7

I forgot what () means tho... is that just a null set? or does it' respresent 0?

Edit: Actually.. {} represents a null set... but... does () represent anything? I'm too far gone from algebra to remember. If ()=0 then the above is your answer.

 

[NotMSRP]

2+2+2+2
1+1+1+1+1+1+1+1
6+(1*1)
6+1
7
--------------------
2+2+2+2
(1+1)+(1+1)+(1+1)+(1+1)
(1+1+1)+(1+1+1)+(1+1)
(3)+(3)+(1*1)
3*3*(1)
9*1
9
-------------------------
2+2+2+2
1+1+1+1+1+1+1+1
((5)*2)+1
10+1
11
--------------------------
2+2+2+2
(1+1+1)+(1+1+1+1)+1
3+4+1
(3*4)+1
13
---------------------------
2+2+2+2
1+1+1+1+1+1+1+1
3*5
15

These are all done with valid steps according to what the OP described.

 

[NotMSRP]

Alternative method to 7:
2+2+2+2
2*(1+1+1+1)
2*(1+1+1)+1
2*3+1
7

 

[nitewulf]

Alternative method to 7:
2+2+2+2
2*(1+1+1+1)
2*(1+1+1)+1
2*3+1
7

no, you can't just take out the "1" from the parenthesis. furthermore i dont think breaking down a 2 into sum of 1s is within the rules.

 

[Nerevar]

2+2+2+2
1+1+1+1+1+1+1+1
6+(1*1)
6+1
7
--------------------
2+2+2+2
(1+1)+(1+1)+(1+1)+(1+1)
(1+1+1)+(1+1+1)+(1+1)
(3)+(3)+(1*1)
3*3*(1)
9*1
9
-------------------------
2+2+2+2
1+1+1+1+1+1+1+1
((5)*2)+1
10+1
11
--------------------------
2+2+2+2
(1+1+1)+(1+1+1+1)+1
3+4+1
(3*4)+1
13
---------------------------
2+2+2+2
1+1+1+1+1+1+1+1
3*5
15

These are all done with valid steps according to what the OP described.

Alternative method to 7:
2+2+2+2
2*(1+1+1+1)
2*(1+1+1)+1
2*3+1
7

wha? All of those involve you doing something wrong. They're not valid uses of algebra at all.

 

[Outlaw]

There is no answer. We've all been fooled :lol :lol This is just to get kids stuck trying to figure out the answer and find out how creative some of them are = ) This is a good way in making math fun...or in some cases more fustrating. I can imagine the smartest kid face when he can't figure out the answer. :lol :lol :lol

 

[Raistlin]

Check this out:

2^() + (2+2+2) = 7

I forgot what () means tho... is that just a null set? or does it' respresent 0?

Edit: Actually.. {} represents a null set... but... does () represent anything? I'm too far gone from algebra to remember. If ()=0 then the above is your answer.

We have a winner imo.

This works in a calculator ... so why the fuck not?

 

[Jenov]

The man who used .2 is the man who solved the riddle.

 

[Raistlin]

No he didn't, as decimals are not allowed.



2^() + (2+2+2) should be allowed since the '^' is not actually needed (I just use it since I can't display superscripts).

 

[Shompola]

We have a winner imo.

This works in a calculator ... so why the fuck not?

Well ^ is an operator and it isnt allowed to be used here.

Btw I checked similiar problems at mathforum and it seemed that the people who answered this kind of problems assumed that you could either use 22 as two twos or use .2 as a single 2...

 

[Jenov]

Well ^ is an operator and it isnt allowed to be used here.

Btw I checked similiar problems at mathforum and it seemed that the people who answered this kind of problems assumed that you could either use 22 as two twos or use .2 as a single 2...

:bow

 

[NotMSRP]

There is no equals sign so I do not have to perserve any properties on both sides of the equation.

0 is done like this:
2+2+2+2 //Could be anything else like 2/2-2*2
(2+2)+(2+2)
(2/2)+(2/2)
(2/2)-(2/2)


22 is not allowed because it requires the concatenation operator.

 

[Raistlin]

Well ^ is an operator and it isnt allowed to be used here.

Btw I checked similiar problems at mathforum and it seemed that the people who answered this kind of problems assumed that you could either use 22 as two twos or use .2 as a single 2...

Did you read the rest of my statement? The '^' is not needed, I just can't display superscripts.

Your logic fails you. How can you argue the '^' is not allowed, but then the '.' is? Neither were represented in the initial list of operators. However, since the '^' is not actually needed, there is at least some merit to the use of exponentiation. In reality, whether the '^' is implied or not, it isn't even conventionally used (outside of programmatic syntax) for diagraming exponentiation - superscripting is.



Some other solutions based on the same premise:

-(2^()) + (2 * 2 * 2)

-(2^()) + (2 * (2^2))

 

[Jenov]

Did you read the rest of my statement? The '^' is not needed, I just can't display superscripts.

Your logic fails you. How can you argue the '^' is not allowed, but then the '.' is?




Some other solutions based on the same premise:

-(2^()) + (2 * 2 * 2)

-(2^()) + (2 * (2^2))

Because .2 is the last simplification. anything with a "^" means it can be simplified more.

 

[Shompola]

notmsrp, I think it is fascinating.. but I have no idea what you are doing sorry...
It reminds me of modulo calculation and that would actually be a good start to see if you can factorize your number to prime numbers. Is that what you are doing here?

for example 7 modulo 2 = 1
i.e. 7 = 2*3+1
And if we transform it to only prime numbers 7 = 2 + 2 + 2 + 1 = 2 + 2+ 2 + 2/2
five twos.. failure.

 

[Raistlin]

Because .2 is the last simplification. anything with a "^" means it can be simplified more.

I'm not following your logic.

What do the comfines of this problem have to do with simplification?

2/2 + 2/2 = 2 is obviously not a simplification.



The rules governing this exercise are pretty clear cut. It states the operators that are available, and the decimal point is not one of them.

Superscripting, the most commonly used syntax for exponentiation, should be allowable within the defined confines as the operator is not actually needed.

 

[Raistlin]

notmsrp, I think it is fascinating.. but I have no idea what you are doing sorry...
It reminds me of modulo calculation and that would actually be a good start to see if you can factorize your number to prime numbers. Is that what you are doing here?

for example 7 modulo 2 = 1
i.e. 7 = 2*3+1
And if we transform it to only prime numbers 7 = 2 + 2 + 2 + 1 = 2 + 2+ 2 + 2/2
five twos.. failure.

Yeah ... I can think of several ways with five 2's :/

 

[Shompola]

Did you read the rest of my statement? The '^' is not needed, I just can't display superscripts.

Your logic fails you. How can you argue the '^' is not allowed, but then the '.' is? Neither were represented in the initial list of operators. However, since the '^' is not actually needed, there is at least some merit to the use of exponentiation. In reality, whether the '^' is implied or not, it isn't even conventionally used (outside of programmatic syntax) for diagraming exponentiation - superscripting is.



Some other solutions based on the same premise:

-(2^()) + (2 * 2 * 2)

-(2^()) + (2 * (2^2))

Yah I did. But as said ^ is an operator.
'.' is not an operator, it is a decimal sign. and what does () mean? the empty set? doing operations on an empty set would not give an output. And if its just parentesis then it is just wrong. I have never seen 2^() in math.
And what is super scripting?

 

[Raistlin]

Yah I did. But as said ^ is an operator.
'.' is not an operator, it is a decimal sign. and what does () mean? the empty set? doing operations on an empty set would not give an output. And if its just parentesis then it is just wrong. I have never seen 2^() in math.
And what is super scripting?

When you learned exponents, was it with the ^ operator? It certainly wasn't for me. Again, I only used it because I can't display superscripts.

Why would it matter whether its implied or not? There was nothing in the rules that says you can't use standard mathetimatical conventions.

If you want to be anal, the decimal is an operator as well. It is a shortcut used in base 10 mathemics for x / (10 * decimal position).

You can either look at it as an complex operator, or as a shortcut (replacement). Either way, it violates the rules. As an operator, it is not part of the list. As a shortcut, it uses a 10 and a whatever the position is.



As far as 2^() in math ... obviously it isn't normally used, but that isn't the point of this exercise. Plug it into a calculator - it works, hack or not. Parenthesis are not used as set operators, they are just used for precedence. However, using them implies a value exists - so if none is stated, it must be a value equivelent to 0.



If you disagree with my logic ... fine, you can argue away the use of exponents. However, you can't do that and then say a decimal is allowed.

 

[Shompola]

You do have a good point on decimals.

 

[NotMSRP]

I am not using modular arithmetic cause it is an operation and one not allowed according the rules described.

. is indeed a shortcut for a more formal form consisting of various operation(s).
Rational numbers can be expressed as a / b where a and b are the normal integers you are used to.
Irrational numbers are tricker. Some can be represented as in exponentiation as in 2^1/2, the square root of 2. Some others are in infinite summations.

() by itself has no meaning so 2^() has no meaning.