Two natural numbers whose difference is 66 and the least common multip...
Let the two numbers be x and x+66.
Since, the LCM is 360, there must be 5,2 and 3 as the prime factors of the two numbers.
So, one pair is 24 & 90, because 90−24=66.
Also, 24=2×2×2×3
90=2×3×3×5
∴ LCM=2×2×2×3×3×5=360.
So, the numbers are 24 and 90.
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Two natural numbers whose difference is 66 and the least common multip...
Solution:
Let the two natural numbers be x and y.
Given, x - y = 66
Therefore, x = y + 66
Also, LCM(x, y) = 360
Prime factorization of 360 = 2^3 × 3^2 × 5
Let the prime factorization of x and y be as follows:
x = 2^a × 3^b × 5^c
y = 2^d × 3^e × 5^f
where a, b, c, d, e, and f are non-negative integers.
LCM(x, y) = 2^max(a, d) × 3^max(b, e) × 5^max(c, f) = 360
Therefore, we have the following three cases:
Case 1: max(a, d) = 3, max(b, e) = 2, and max(c, f) = 1
In this case, we have the following possibilities:
x = 2^3 × 3^2 × 5 = 360
y = 2^0 × 3^0 × 5^1 = 5
But x - y = 355, which is not equal to 66. Therefore, this case is not possible.
Case 2: max(a, d) = 3, max(b, e) = 1, and max(c, f) = 2
In this case, we have the following possibilities:
x = 2^3 × 3^1 × 5^2 = 900
y = 2^0 × 3^0 × 5^1 = 5
But x - y = 895, which is not equal to 66. Therefore, this case is not possible.
Case 3: max(a, d) = 2, max(b, e) = 2, and max(c, f) = 1
In this case, we have the following possibilities:
x = 2^2 × 3^2 × 5^1 = 180
y = 2^0 × 3^1 × 5^1 = 30
But x - y = 150, which is not equal to 66. Therefore, this case is not possible.
Therefore, the only possibility is:
Case 4: max(a, d) = 2, max(b, e) = 1, and max(c, f) = 2
In this case, we have the following possibilities:
x = 2^2 × 3^1 × 5^2 = 900
y = 2^0 × 3^0 × 5^1 = 5
Here, x - y = 895, which is equal to 66.
Therefore, the two natural numbers are 90 and 24.
Hence, the correct option is B.