SOLUTION: Find the number of ordered pairs (m,n) of integers that satisfy mn = 3m + 3n + 17.

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Question 1209235: Find the number of ordered pairs (m,n) of integers that satisfy
mn = 3m + 3n + 17.

Found 4 solutions by greenestamps, AnlytcPhil, ikleyn, math_tutor2020:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!




Solve the equation for one variable in terms of the other:





Perform the indicated division and express the result as quotient and remainder:




In that last equation, m and 3 are integers, so must be an integer.

The number of ordered pair solutions is the number of integer factors of 26, which is 4.

NOTE: In typical problems like this, we are looking for solutions in positive integers. However, this problem does not specify positive integers; counting positive and negative integers, the number of integer factors of 26 is 8.

So there are 8 ordered pair solutions.

ANSWER: 8

The problem doesn't ask us to find the solutions, but we can do so to verify that there are 4 pairs of solutions. Note that the expression is symmetrical in m and n, so if (a,b) is a solution the (b,a) will be a solution. So to find the 8 solutions we only need to find 4 of them and switch the order of the two numbers to get the other solutions.

   n-3  n  m=3+26/(n-3)   solutions (m,n)
  ---------------------------------------
    1   4   3+26/1 = 29     (4,29) and (29,4)
    2   5   3+26/2 = 16     (5,16) and (16,5)
   -1   2   3+26/-1 = -23   (2,-23) and (-23,2)
   -2   1   3+26/-2 = -10   (1,-10) and (-10,1)



Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!






Long division:

        3
n-3)3n+17
    3n- 9
       26



n-3 must divide evenly into 26, so n-3 is either ±1,±2,±13, or ±26

That's 8 possibilities, so we know the answer is 8. To check, we find them:

1. if n-3 = -26, then n = -23, and m = 3+26/(-23-3) = 2, ordered pair (2,-23).
2. if n-3 = -13, then n = -10, and m = 3+26/(-10-3) = 1, ordered pair (1,-10).
3. if n-3 = -2, then n = 1, and m = 3+26/(1-3) = -10, ordered pair (-10,1).
4. if n-3 = -1, then n = 2, and m = 3+26/(2-3) = -23, ordered pair (-23,2).
5. if n-3 = 1, then n = 4, and m = 3+26/(4-3) = 29, ordered pair (29,4).
6. if n-3 = 2, then n = 5, and m = 3+26/(5-3) = 16, ordered pair (16,5).
7. if n-3 = 13, then n = 16, and m = 3+26/(16-3) = 5, ordered pair (5,16).
8. if n-3 = 26, then n = 29, and m = 3+26/(29-3) = 4, ordered pair (4,29).


So indeed, there are 8 such ordered pairs.

Here is a plot of all 8.  See how nicely they fit on that curve.

Can you guess what the equation of that curve is?


Edwin

Answer by ikleyn(52779)   (Show Source): You can put this solution on YOUR website!
.
Find the number of ordered pairs (m,n) of integers that satisfy
mn = 3m + 3n + 17.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Transform the given equation this way

    mn - 3m - 3n + 9 = 26,

    (n-3)*(m-3) = 26.


So, n-3 and m-3 are divisors of 26.


There are 4 different decompositions of 26 into the product of positive integer numbers

    8 = 1*26 = 2*13 = 13*2 = 26*1


It gives 4 different pairs (m,n).         // Since the problem does not ask to find them,
                                             I will not find them.



There are 4 different decompositions of 26 into the product of negative integer numbers

    26 = (-1)*(-26) = (-2)*(-13) = (-13)*(-2) = (-26)*(-1).


It gives 4 other different pairs (m,n).   // Since the problem does not ask to find them,
                                             I will not find them.


In all, there are 4+4 = 8 different pairs of integer numbers, satisfying the given equation.

Solved.



Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Answer: 8

Work Shown

mn = 3m+3n+17
mn-3m-3n = 17
mn-3m-3n+9 = 17+9
(mn-3m)+(-3n+9) = 26
m(n-3)-3(n-3) = 26
(m-3)(n-3) = 26

n = 26 = 2^1*13^1 = p^a*q^b shows that a = 1, b = 1
d(n) = number of positive divisors of n
n = p^a*q^b
d(n) = (1+a)*(1+b)
d(26) = (1+1)*(1+1)
d(26) = 4
There are 4 positive divisors of 26 and they are: 1, 2, 13, 26.
When including the negative divisors we get 8 total divisors:
1, 2, 13, 26,
-1, -2, -13, -26

So there must be 8 ordered pairs (m,n) that make (m-3)(n-3) = 26 true and that make m*n = 3m+3n+17 true as well; where m,n are integers.
If m,n were allowed to be decimal numbers, then there would be infinitely many ordered pair solutions.

-----------------------

Extra info:
If you want to find each ordered pair, then set the factors m-3 and n-3 equal to the factors of 26.

Some examples:
m-3 = 1 and n-3 = 26 lead to (m,n) = (4,29)
m-3 = 2 and n-3 = 13 lead to (m,n) = (5,16)
m-3 = 13 and n-3 = 2 lead to (m,n) = (16,5)

Here are all 8 integer solutions
(-23,2), (-10,1), (1,-10), (2,-23),
(4,29), (5,16), (16,5), (29,4)
They are sorted from smallest x coordinate to the largest x coordinate.
We have symmetry going on since (5,16) swaps to (16,5) as one example.

Side notes:

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