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Paul
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PostPosted: Fri Dec 16, 2005 2:02 pm 
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sirclucky wrote:
Comeon guys. I got two entries for the first round, and one incorrent entry for the second round. im not gonna make challenges if no one will respond.


Hey! I got a mention, that rocks. You can't figure it without having one answer.
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sirclucky
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PostPosted: Fri Dec 16, 2005 2:08 pm 
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Even with both you are Paul, that would still only be two, tested.

To solve the problem, call the five answers A B C D and E

Question One: This is the answer to question two plus 3. (A=B+3)
Question Two: This is twice the answer to question one minus the answer to question four. (B=2*A-D)
Question Three: This is the total number of questions which have an odd answer times the answer to question five. ( C=odd*E)
Question Four: This is the answer to question one plus half the answer to question three. (A+1/2*C)
Question Five: This is twice the answer to question two minus 6. (E=2*B-6)

B=2*A-D
A=B+3
2*A=2*B+6
B=2*B+6-D

B+6=D
B+3=A
B=B
2*B-6=E

D=A+1/2*C
B+6=B+3+1/2*C
3=1/2*C
6=C

So either there is one odd number, and E is six. There are two odd numbers, and E is three, or there is one odd number, and E is two.
E is the difference of two even numbers, and thus must be even.
So either E is 6 or E is two.
If 2*B-6=2
2*B=8
B=4
D=10
A=7
C=6
E=2

There is only one odd number. Hence C would accually be 2*1=2. So E cannot be two.

If E is six then
2*B-6=6
2*B=12
B=6

A=6+3=9
B=6
C=6
D=+2+=12
E=6

Doubling checking that these all work.
A=9=B+3=6+3
B=2*A-D=2*9-12=18-12=6
C=*odds**E=1*6=6
D=A+1/2*C=9+6/2=9+3=12
E=2*B-6=2*6-6=12-6=6

Yay!

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Uncle Xyzzy
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PostPosted: Fri Dec 16, 2005 2:59 pm 
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sirclucky wrote:
Even with both you are Paul, that would still only be two, tested.

To solve the problem, call the five answers A B C D and E

Question One: This is the answer to question two plus 3. (A=B+3)
Question Two: This is twice the answer to question one minus the answer to question four. (B=2*A-D)
Question Three: This is the total number of questions which have an odd answer times the answer to question five. ( C=odd*E)
Question Four: This is the answer to question one plus half the answer to question three. (A+1/2*C)
Question Five: This is twice the answer to question two minus 6. (E=2*B-6)

B=2*A-D
A=B+3
2*A=2*B+6
B=2*B+6-D

B+6=D
B+3=A
B=B
2*B-6=E

D=A+1/2*C
B+6=B+3+1/2*C
3=1/2*C
6=C

So either there is one odd number, and E is six. There are two odd numbers, and E is three, or there is one odd number, and E is two.
E is the difference of two even numbers, and thus must be even.
So either E is 6 or E is two.
If 2*B-6=2
2*B=8
B=4
D=10
A=7
C=6
E=2

There is only one odd number. Hence C would accually be 2*1=2. So E cannot be two.

If E is six then
2*B-6=6
2*B=12
B=6

A=6+3=9
B=6
C=6
D=+2+=12
E=6

Doubling checking that these all work.
A=9=B+3=6+3
B=2*A-D=2*9-12=18-12=6
C=*odds**E=1*6=6
D=A+1/2*C=9+6/2=9+3=12
E=2*B-6=2*6-6=12-6=6

Yay!


Wow. I need to show this to Justin.

Who is Justin, you ask? A genius.

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Uncle Xyzzy
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PostPosted: Fri Dec 16, 2005 3:01 pm 
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sirclucky wrote:
Even with both you are Paul, that would still only be two, tested.

To solve the problem, call the five answers A B C D and E

Question One: This is the answer to question two plus 3. (A=B+3)
Question Two: This is twice the answer to question one minus the answer to question four. (B=2*A-D)
Question Three: This is the total number of questions which have an odd answer times the answer to question five. ( C=odd*E)
Question Four: This is the answer to question one plus half the answer to question three. (A+1/2*C)
Question Five: This is twice the answer to question two minus 6. (E=2*B-6)

B=2*A-D
A=B+3
2*A=2*B+6
B=2*B+6-D

B+6=D
B+3=A
B=B
2*B-6=E

D=A+1/2*C
B+6=B+3+1/2*C
3=1/2*C
6=C

So either there is one odd number, and E is six. There are two odd numbers, and E is three, or there is one odd number, and E is two.
E is the difference of two even numbers, and thus must be even.
So either E is 6 or E is two.
If 2*B-6=2
2*B=8
B=4
D=10
A=7
C=6
E=2

There is only one odd number. Hence C would accually be 2*1=2. So E cannot be two.

If E is six then
2*B-6=6
2*B=12
B=6

A=6+3=9
B=6
C=6
D=+2+=12
E=6

Doubling checking that these all work.
A=9=B+3=6+3
B=2*A-D=2*9-12=18-12=6
C=*odds**E=1*6=6
D=A+1/2*C=9+6/2=9+3=12
E=2*B-6=2*6-6=12-6=6

Yay!


Wow. I need to show this to Justin.

Who is Justin, you ask? A genius.

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sirclucky
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PostPosted: Fri Dec 16, 2005 5:04 pm 
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Heh. Sorry if you found it hard. Math comes naturally to me, and there was nothing more than algebra in that puzzle, so I guess im just not a good judge at whats hard and whats not =P

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Uncle Xyzzy
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PostPosted: Sat Dec 17, 2005 11:58 pm 
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sirclucky wrote:
Heh. Sorry if you found it hard. Math comes naturally to me, and there was nothing more than algebra in that puzzle, so I guess im just not a good judge at whats hard and whats not =P


Oh, I just had trouble with the fact that this problem produces an infinite series. Many college students don't know how to do it, and i only know how to express it. Because several variables had the capability of equaling themselves plus some, it can produce an infinite series, ie:

f(X)=2^f(X)+f(X) for any X.

Agh.

Except yours is more linear, which makes it a slight bit easier.

Next challenge?

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paperfacesX022
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PostPosted: Sun Dec 18, 2005 6:05 pm 
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And I thought I was good at math . . .

*just a note*

I'm in 7th grade. I tried this challenge, but I just couldn't get it. Sorry I didn't enter this one, Clucky.

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