What Is the Correct Calculation for 643 + 364 Using Defined Functions?

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Without a clear definition, it's impossible to solve this problem. Additionally, the given attempt at a solution is not helpful in understanding the problem. In summary, the question is not giving enough information to be solved and the provided attempt at a solution is not helpful.
  • #1
Greychu
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Homework Statement


For positive integers m, k, and n , let mkn be defined as mkn = kmn , where [itex] k\frac {m}{n} [/itex] is a mixed fraction. What is the value of 643 + 364 ?

Homework Equations


I attempt the other few similar questions where the solution are as follow
832 + 382 = [itex] \frac {169}{24} [/itex]
641 + 164 = [itex] \frac {85}{6} [/itex]

The answer for this question.
643 + 364 = [itex] \frac {21}{2} [/itex]

But the problem is I have no idea how to get the answer.

The Attempt at a Solution


Based on the problem, I can see that

643 + 364 = 463 + 634
= 463 + 364
= 643 + 634
or more simply

643 = 463 and 364 = 634

Notice that 463 and 364 are in reverse position with each other.

Since it mention that [itex] m\frac {k}{n} [/itex] is a mixed fraction, I suspect that
643 + 364 = 643 + 634

I try to relate 643 = [itex] 6 \frac {4}{3} [/itex] and 364 = [itex] 3 \frac {6}{4} [/itex]
but they are not the same for 463 and 634 if expressed in term of mixed fraction

Even attempt the way:
643 = 463 → (24)3 and 364 = 634 → (18)4 by multiply the first 2 numbers and
643 = 463 = (10)3 and 364 = 634 → (9)4 by adding the first 2 numbers.

Not sure if mkn = kmn = [itex] k\frac {m}{n} [/itex]

Please try to give some hints about how to see this problem.
 
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  • #2
Greychu said:

Homework Statement


For positive integers m, k, and n , let mkn be defined as mkn = kmn , where [itex] k\frac {m}{n} [/itex] is a mixed fraction. What is the value of 643 + 364 ?
I don't get it. The sentence above is defining mkn as kmn. How does ##k\frac m n## tie into either of the expressions shown earlier?
If there is an actual definition here, I'm not seeing it.
Greychu said:

Homework Equations


I attempt the other few similar questions where the solution are as follow
832 + 382 = [itex] \frac {169}{24} [/itex]
641 + 164 = [itex] \frac {85}{6} [/itex]

The answer for this question.
643 + 364 = [itex] \frac {21}{2} [/itex]

But the problem is I have no idea how to get the answer.

The Attempt at a Solution


Based on the problem, I can see that

643 + 364 = 463 + 634
= 463 + 364
= 643 + 634
or more simply

643 = 463 and 364 = 634

Notice that 463 and 364 are in reverse position with each other.

Since it mention that [itex] m\frac {k}{n} [/itex] is a mixed fraction, I suspect that
643 + 364 = 643 + 634

I try to relate 643 = [itex] 6 \frac {4}{3} [/itex] and 364 = [itex] 3 \frac {6}{4} [/itex]
but they are not the same for 463 and 634 if expressed in term of mixed fraction

Even attempt the way:
643 = 463 → (24)3 and 364 = 634 → (18)4 by multiply the first 2 numbers and
643 = 463 = (10)3 and 364 = 634 → (9)4 by adding the first 2 numbers.

Not sure if mkn = kmn = [itex] k\frac {m}{n} [/itex]

Please try to give some hints about how to see this problem.
 
  • #5
Is the question not giving enough information to be solved?
 
  • #6
Greychu said:
Is the question not giving enough information to be solved?
In my opinion, and apparently that of haruspex, no.
 
  • #7
Greychu said:

Homework Statement


For positive integers m, k, and n , let mkn be defined as mkn = kmn , where [itex] k\frac {m}{n} [/itex] is a mixed fraction. What is the value of 643 + 364 ?

Homework Equations


I attempt the other few similar questions where the solution are as follow
832 + 382 = [itex] \frac {169}{24} [/itex]
641 + 164 = [itex] \frac {85}{6} [/itex]

The answer for this question.
643 + 364 = [itex] \frac {21}{2} [/itex]

But the problem is I have no idea how to get the answer.

The Attempt at a Solution


Based on the problem, I can see that

643 + 364 = 463 + 634
= 463 + 364
= 643 + 634
or more simply

643 = 463 and 364 = 634

Notice that 463 and 364 are in reverse position with each other.

Since it mention that [itex] m\frac {k}{n} [/itex] is a mixed fraction, I suspect that
643 + 364 = 643 + 634

I try to relate 643 = [itex] 6 \frac {4}{3} [/itex] and 364 = [itex] 3 \frac {6}{4} [/itex]
but they are not the same for 463 and 634 if expressed in term of mixed fraction

Even attempt the way:
643 = 463 → (24)3 and 364 = 634 → (18)4 by multiply the first 2 numbers and
643 = 463 = (10)3 and 364 = 634 → (9)4 by adding the first 2 numbers.

Not sure if mkn = kmn = [itex] k\frac {m}{n} [/itex]

Please try to give some hints about how to see this problem.

Your definition (or, rather, the definition in your source) is vague and almost useless. Do you mean
[tex] m\underline{k}n = m\, \frac{k}{n} \;(\text{which} \: = m + \frac{k}{n} = \frac{mn + k}{n}) [/tex]
or do you mean
[tex] m\underline{k}n = m \times \frac{k}{n} \:(\text{which} \: = \frac{mk}{n} )? [/tex]

Note that ##a\underline{b}c = b\underline{a}c## for the second definition of ##a\underline{b}c##, etc., but not for the first one.

The people who set this question should be ashamed, unless they have already defined the notation ##a \frac{b}{c}## somewhere else in the paper.
 
  • #8
Here is an image of the problem. (It looks as if the website randomizes the the numbers.) It a real puzzle as to what that operation is.
?temp_hash=e1642ed97ddabf9570e6574d9ace7c16.png

The correct answer is (C) . I have no idea why that's the case.

To view the solution seems to require signing up for a fee. (I'm not curious enough & too cheap for that.) The website is Brilliant.org

SammyS
 

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  • #9
That "definition" is meaningless since it defines "[itex]m\underline{k}n[/itex]" as "[itex]k\underline{m}n[/itex]" which has not been defined!

This is like defining a number to be "whacky" by "x is 'whacky' if and only if x/2 is 'whacky'. The definition is circular.
 
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  • #10
alright, I think I have found the way to solve it
Let take 643 + 364
= 463 + 634

I got the answer by performing this:

[tex](\frac {(24)(24)}{(18)(16)} + \frac {9}{18})+8[/tex] = [tex] \frac {21}{2}[/tex]

Notice that first 2 term 6 x 4 = 4 x 6 = 24
final 2 term 6 x 3 = 18 and 4 x 4 = 16 respectively
The fraction consists of
[tex] \frac {3 x 3}{3 x 6}[/tex] = [tex] \frac {9}{18}[/tex]
the 8 = 4 + 4 is the addition of last digit of the last number.

I try this and it applies to all the similar questions. I still cannot find a logical reason how it leads to this, so if you got any insights, can let me know.
 
  • #11
Five people have responded in this thread, and none of them can make any sense out of this problem. Since the problem is so poorly defined, the best thing to do would be to forget it, and move on to other questions.
 
  • #12
Mark44 said:
Five people have responded in this thread, and none of them can make any sense out of this problem. Since the problem is so poorly defined, the best thing to do would be to forget it, and move on to other questions.
As we suspected, there was some error in the statement of the problem. - - a typo or some major definition left out. I just looked it up again. Here's an image from today.
PhysForum_NewlyDefFunc_2.PNG

So, it's not too interesting a problem after all.

This doesn't seem to work for some of the examples cited yesterday. ... or do I have brain fade?
 
Last edited:
  • #13
I can't make any sense of the questions/answers in your post but when i tried the link the most similar looking question i got was
For positive integers ##m,k,n## and , let ##m\mathbf kn## be defined as ##m\mathbf kn = \mathbf k \frac{m}{n}##, where ##\mathbf k \frac{m}{n}## is a mixed fraction. What is the value of ##6\mathbf 21 + 1\mathbf 26##
so we get
##2\frac{6}{1} + 2\frac{1}{6} = 8 + \frac{13}{3} = \frac{48+13}{6} = \frac{61}{6}##
which gives me correct answer according to that site. However for the answers in your question i get the wrong answer so maybe I found the wrong question.

Edit: I guess the problem was the question had an error in it when you tried it before
 

1. What are newly defined functions?

Newly defined functions are functions that have been created by the programmer to perform a specific task or set of tasks. They are not built-in to the programming language and must be defined by the user.

2. Why are newly defined functions useful?

Newly defined functions allow for code reusability, making it easier to write and maintain complex programs. They also help to break down a large program into smaller, more manageable parts.

3. How do you define a new function?

To define a new function, you must first choose a name for the function and specify the parameters it will take (if any). Then, you write the code that the function will execute when called. The function must also have a return statement to specify what value it will return.

4. Can newly defined functions be modified?

Yes, newly defined functions can be modified by the programmer. This allows for flexibility and the ability to improve the function over time.

5. How do you call a newly defined function?

To call a newly defined function, you simply use the function's name followed by parentheses, with any necessary arguments inside the parentheses. This will execute the code within the function and return the specified value.

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