Question 1165024: John thinks of two positive integers. He multiplies them together and then
subtracts each of the integers from the product, with a result of 35. Find
all possible pairs of numbers he could have chosen.
Found 3 solutions by greenestamps, Edwin McCravy, ikleyn:
Answer by greenestamps(13208)   (Show Source):
You can put this solution on YOUR website!
Let the integers be x and y. Then
This is a single equation in two variables, with both variables being integers. To find all the solutions, solve the equation for one variable in terms of the other and use the fact that the variables are integers to find the solutions.
You could use this form of the equation to find values of x that make y an integer. However, it is easier to perform the division on the right to express it as an integer plus a remainder.
In this form, we can see that y will be an integer whenever (x-1) is a factor of 36.
x-1 x y = 1+36/(x-1) (x,y)
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1 2 1+36/1 = 37 (2,37)
2 3 1+36/2 = 19 (3,19)
3 4 1+36/3 = 13 (4,13)
4 5 1+36/4 = 10 (5,10)
6 7 1+36/6 = 7 (7,7)
9 10 1+36/9 = 5 (10,5)
12 13 1+36/12 = 4 (13,4)
18 19 1+36/18 = 3 (19,3)
36 37 1+36/36 = 2 (37,2)
It should be clear that the expression is symmetric in x and y, so -- as shown in the table of solutions -- for each solution (a,b), (b,a) will also be a solution.
There are 9 solutions as ordered pairs; if order does not matter, there are 5 solutions.
Answer by Edwin McCravy(20063)  (Show Source):
You can put this solution on YOUR website! John thinks of two positive integers.
Suppose he think of positive integers A and B
He multiplies them together
He multiplies them together and gets AB
and then subtracts each of the integers from the product,
He subtracts A and gets AB-A
He then subtracts B and gets AB-A-B
with a result of 35.
So
AB-A-B = 35
A(B-1)-B = 35
A(B-1) = B+35
So we can pick B-1 as any allowable divisor of 36,
The divisors of 36 are 1,2,3,4,6,9,12,18,36
So B-1 can only be one of those.
which means
B = 2,3,4,5,7,10,13,19,37
Then substituting each in
the corresponding values for A are
A = 37,19,13,10,7,5,4,3,2
So we could have
{A,B} = {2,37), (3,19), (4,13}, (5,10}, {7,7}
Edwin
Answer by ikleyn(52875)  (Show Source):
You can put this solution on YOUR website! .
I came to solve the problem in a way as it SHOULD be done.
Let x and y be two positive integers John thinks about.
Then from the condition, you have this equation
xy - x - y = 35.
Add 1 (one) to both side. You will get
xy - x - y + 1 = 36.
It is equivalent to
(x-1)*(y-1) = 36
Thus the integer number 36 is presented as the product of integer positive factors (x-1) and (y-1).
The possible factors of the number 36 are
1, 2, 3, 4, 6, 9, 12, 18, 36.
So, (x-1) may have these values 1, 2, 3, 4, 6, 9, 12, 18, 36
(and, correspondingly, (y-1) has the values of the same list in reversed order).
So, all possible values for numbers x and y are these pairs
(x,y) = (2,37), (3,19), (4,13), (7,7), (10,5), (13,4), (19,3) and (37,2). ANSWER
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Solved. // In a way as it SHOULD be.
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