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Math sequence

  • Thread starter daelin
  • Start date
  • #1
5
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help me i dont know where i must to posting this task

if
8+4=12
7+2=9
2+3=6
6+5=10



9+5 = ?
 

Answers and Replies

  • #2
Mentallic
Homework Helper
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I don't understand...

2+3=5 and 6+5=11 but you've shown otherwise, and I don't understand how we are now going to conclude what 9+5 is equal to based on this wrong maths.
 
  • #3
5
0
I don't understand...

2+3=5 and 6+5=11 but you've shown otherwise, and I don't understand how we are now going to conclude what 9+5 is equal to based on this wrong maths.
figured number

13+7=5
15+6=9
10+7=11
8+8=11

this for next example

please help :D
 
  • #4
Mentallic
Homework Helper
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I still don't understand. Can you give me an example where you have the solution and explain why the answer is as it is?

Right now I'm just seeing a few additions that a kid failing kindergarten tried to attempt.
 
  • #5
311
1
Is this a riddle? Or are you asking for help on answering a question?
 
  • #6
33,484
5,174
help me i dont know where i must to posting this task

if
8+4=12
7+2=9
2+3=6
6+5=10

9+5 = ?
Much of this doesn't make any sense. 2 + 3 = 5, not 6, and 6 + 5 = 11, not 10.


figured number

13+7=5
15+6=9
10+7=11
8+8=11

this for next example
And none of the above makes any sense.
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,833
955
help me i dont know where i must to posting this task

if
8+4=12
7+2=9
2+3=6
6+5=10



9+5 = ?
I assume that this is some algebraic structure in which the digits do NOT refer to the usual integers and/or "+" is not ordinary "add". But the difficulty is, what can we take to be true? Are we to assume this operation is commutative? Associative?

Assuming that it is associative, 9+ 5= (7+ 2)+ 5 but I don't see where to go form here.
 
  • #8
gb7nash
Homework Helper
805
1
help me i dont know where i must to posting this task

if
8+4=12
7+2=9
2+3=6
6+5=10



9+5 = ?
I'm going to throw in a wild guess and say 14.
 
  • #9
2,967
5
help me i dont know where i must to posting this task

if
8+4=12
7+2=9
2+3=6
6+5=10



9+5 = ?
If each symbol stands for a particular, but different true decimal digit and '+' is a sign for regular addition, then the above conditions do not have a solution. I have check all 151,200 possibilities by brute force.
 
  • #10
Mentallic
Homework Helper
3,798
94
If each symbol stands for a particular, but different true decimal digit and '+' is a sign for regular addition, then the above conditions do not have a solution. I have check all 151,200 possibilities by brute force.
That's what I was thinking.

Essentially we would be solving

a+b=c
d+e=f
e+g=h
...
etc.

Without being given any further information as to how these symbols interact with each other.


But I have a totally new theory, the OP is just trolling us.
 
  • #11
33,484
5,174
But I have a totally new theory, the OP is just trolling us.
That thought crossed my mind, too.
 
  • #12
5
0
That's what I was thinking.

Essentially we would be solving

a+b=c
d+e=f
e+g=h
...
etc.

Without being given any further information as to how these symbols interact with each other.


But I have a totally new theory, the OP is just trolling us.
my teacher just say

"do not be fooled pattern placement sequence similarities.
without divisors and a deduction"
 
  • #13
gb7nash
Homework Helper
805
1
"do not be fooled pattern placement sequence similarities.
without divisors and a deduction"
Clarify? This makes no sense to me. I still think 14's a good answer.
 
  • #14
5
0
there are only 2 of mathematical symbols
x nd +
 
  • #15
47
0
Well, I spent two hours thinking about the problem, took a few hours off, then continued for another hour but I'm still stumped. Is there any more to the sequence?

Z
 
  • #16
47
0
I got a new approach to this problem.

If:
8+4 = 12
3+5 = 17
9+4 = 2
6+5 = 11

Then 9+5 = somebody failed kindergarten.

No offense to you or anything, because this is a very difficult logic problem; whether or not it is actually solvable I don't know. But, I'm taking the easy way out.
 
  • #17
Mentallic
Homework Helper
3,798
94
Well, I spent two hours thinking about the problem, took a few hours off, then continued for another hour but I'm still stumped. Is there any more to the sequence?

Z
Then the troll wins! :yuck:
 
  • #18
If each symbol stands for a particular, but different true decimal digit and '+' is a sign for regular addition, then the above conditions do not have a solution. I have check all 151,200 possibilities by brute force.
That's what I was thinking.

Essentially we would be solving

a+b=c
d+e=f
e+g=h
...
etc.
I too thought the same thing. How did you come up with 151,200 possibilities btw?
 
  • #19
2,967
5
I too thought the same thing. How did you come up with 151,200 possibilities btw?
help me i dont know where i must to posting this task

if
8+4=12
7+2=9
2+3=6
6+5=10



9+5 = ?
The following digits appear: {0, 1, 2, 3, 4, 5, 6, 7, 8 , 9}, i.e. all ten of them. Let us make the correspondence to variables:

0 <-> a
1 <-> b
2 <-> c
3 <-> d
4 <-> e
5 <-> f
6 <-> g
7 <-> h
8 <-> i
9 <-> j

Then, the above condition may be rewritten as:

i + e = 10*b + c
h + c = j
c + d = g
g + f = 10*b + a


However, we have 4 linear relationship between the 10 of them. So, we may choose 6 of them as known and express the other 4 in terms of them. Since we have to find the result of the operation '9 + 5 <-> j + f', I consider f and j among the 4 'unknowns''. The 'known' variables are:

(a, b, c, d, e, h)

Then, the 'unknowns' (f, g, i, j) are expressed as:

f = a + 10*b - c - d
g = c + d
i = 10*b + c - e
j = c + h

We choose the 'knowns' from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. The order matters. Also, we cannot choose the same value for different symbols. Thus, the number of choices is equal to the number of selections without repetition of 10 elements of class 6:

10*9*8*7*6*5 = 151,200

I made a program (in Mathematica) to go through all possibilities. For each selection, it calculates the knowns. Then, it checks these things:

  1. Whether each 'unknown' is between 0 and 9, inclusively
  2. Whether each 'unknown' is different from ANY of the 'knowns'
  3. Whether each 'known' is different from ALL of the other 'unknowns'

Effectively, this checks whether I had generated a permutation of the ten decimal digits. Any such permutation is a viable option. However, the program did not find a single such selection.
 
  • #20
hmm. I assumed the final question would have the answer in the symbol set and the numbers may be just symbols i.e. two numbers are the same just different symbols. Thanks for the explanation.
 
Last edited:
  • #21
5
0
please make is logic
how can 8 + 8 it should 16 but this is 11
10 + 7 should 17 but 11
 
  • #22
Mentallic
Homework Helper
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Did we ever say that 8+8=10+7=11? No, you said it. How can we make logical sense out of your incorrect maths, or wherever you got this incorrect maths?
 
  • #23
43
0
I think it's some kind of series brain teaser where the numbers follow some pattern, but I have no clue what.

Like for example 2 + 3 = 10 where + means to add the numbers and multiply by the first.

I have no clue what the pattern in the poster's example are though. But maybe someone else can think of an operation that + and = could stand for?

... maybe looking at them this way makes it easier to find a pattern:
8,4,12
7,2,9
2,3,6
6,5,10
 
Last edited:

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