The difference of two positive integers was multiplied by their sum. If possible, find two numbers such that the result is equal to 212.
Result from addition or multiplication?
let x be the larger and y be the smaller
(x-y)(x+y) = 212
212 = 2*106
= 4*53
there are no other factors, since 53 is prime
212 = 2* 106 , so the two numbers are 2 apart and have a sum of 106
dividing 106/2 = 53
so how about 54 and 52
x = 54 , y = 52 and
(x-y)(x+y)
= 2(106) = 212
yup that works.
To find two positive integers whose difference, when multiplied by their sum, gives 212, we can set up an equation.
Let's assume the two positive integers are x and y.
According to the problem, the difference of the two positive integers is multiplied by their sum, equating to 212:
(x - y) * (x + y) = 212
Expanding the equation:
x^2 - y^2 = 212
Now, we need to factorize the left side of the equation:
(x - y) * (x + y) = 212
Looking at the factors of 212, we find that 2 and 106 are factors of 212:
2 * 106 = 212
So, we have:
(x - y) * (x + y) = 2 * 106
Now, we need to determine how to select the values for x and y in order to satisfy this equation.
We have several possible solutions:
1. Setting (x - y) = 2 and (x + y) = 106:
Solving these simultaneous equations, we get:
x = 54 and y = 52
2. Setting (x - y) = 106 and (x + y) = 2:
Solving these simultaneous equations, we get:
x = 54 and y = -52 (However, as per the problem, we need to find positive integers, so this solution is not valid.)
In conclusion, the two positive integers that satisfy the given condition are 54 and 52.