Assume that x is an even natural number and y is a natural number such that y<x. If LCM(x,y) =2100 and GCF(x,y)=3 then find x and y
Investigation:
let x = 30 and y = 24
30 = 2*3*5
24 = 2*2*2*3
GCF = 6 and LCM = 120
Suppose we didn't know our original x and y, but we
know their GCF and their LCM
so we can say: x = 6a, and y = 6b
and xy = 6a(6b) = 36 ab = 120 , such that a and b are relatively prime
ab = 120/36 = 20
since a and b are relatively prime, a = 5, b = 4 , (the only choice)
and we get our original x and y with x = 6*5 = 30 and y = 4*6 = 24
.............................
now follow the same steps for yours:
we know that xy = 2100*3 = 6300
let x = 6a, and y = 6b
36ab = 6300
ab = 175
which two factor of 175 are relatively prime??
how about 25*7
so we let a = 25 ----> x = 6*25 = 150
and let b = 7 -----> y = 6*7 = 42
In my example, the two lines should say:
and xy = 720 , 6a(6b) = 36ab = 720 , such that a and b are relatively prime
ab = 720/36 = 20
(I typed 120 instead of 720)
To find the values of x and y, we need to understand the concepts of LCM (Least Common Multiple) and GCF (Greatest Common Factor).
LCM: The LCM of two or more numbers is the smallest multiple that is evenly divisible by all the numbers. In this case, the LCM of x and y is given as 2100.
GCF: The GCF of two or more numbers is the largest number that divides both numbers evenly. In this case, the GCF of x and y is given as 3.
Since x is an even natural number, it can be represented as x = 2n, where n is a natural number.
Now, let's break down the prime factorization of 2100 to determine the possible values for x and y:
Prime factorization of 2100:
2100 = 2^2 * 3 * 5^2 * 7
Since the GCF of x and y is 3, we know that both x and y must be divisible by 3.
Therefore, we can rewrite the prime factorization of 2100 as:
2100 = (2^2 * 5^2 * 7) * 3
Comparing this with the prime factorization of x = 2n:
2n = (2^m * 5^p * 7^q) * 3
Since x is an even natural number, the exponent of 2 (m) must be greater than or equal to 2. This ensures that x is divisible by 4.
Hence, we can conclude that:
m ≥ 2
Now we can determine the possible values of x and y.
x = 2^m * 3 * (5^p * 7^q)
= 2^2 * 3 * (5^2 * 7)
= 420
The value of x is 420.
To determine the value of y, we divide 2100 by x:
y = 2100 / x
= 2100 / 420
= 5
Therefore, the values of x and y are:
x = 420
y = 5