There are 5 consecutive natural numbers with L.C.M. 60. The product of the first two numbers is equal to the 5th number. What is the sum of the numbers?a)36b)25c)20d)15Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Let n be 1^st natural number.
So numbers be
n, n + 1, n + 2, n + 3, n + 4, n + 5
According to question
► n(n + 1) = n + 4
► n² + n = n + 4
► n² = 4
► n = 2
– 2 cannot be taken as n is natural number.
So n = 2 numbers be 2, 3, 4, 5, 6
Sum of numbers = 2 + 3 + 4 + 5 + 6 = 20.

Answer

Approach:
Let us assume the first number to be n. Then the consecutive numbers will be n+1, n+2, n+3 and n+4.
We know that the L.C.M. of these numbers is 60. Therefore, we can write:
lcm(n,n+1,n+2,n+3,n+4) = 60
Now, let us find the prime factors of 60: 2, 2, 3, 5
We need to find a combination of these factors that will give us the numbers n, n+1, n+2, n+3 and n+4.
- If we include 2 and 3, we can get at most 3 consecutive numbers.
- If we include 5, then we need to have at least one number that is a multiple of 5.
Hence, the only possible combination is:
n = 2 x 3 x 5 = 30
n+1 = 31
n+2 = 32
n+3 = 33
n+4 = 34

Now, we are given that the product of the first two numbers (30 and 31) is equal to the fifth number (34).
Therefore, we can write:
30 x 31 = 34
This is not true, so there is no solution.

Let us assume that the product of the first two numbers is equal to the fourth number (n+3).
Therefore, we can write:
n(n+1) = n+3 x 60
Expanding, we get:
n^2 + n = 60n + 180
n^2 - 59n - 180 = 0
Solving this quadratic equation, we get:
n = 12 or n = 47

If n = 12, then the consecutive numbers are:
12, 13, 14, 15, 16
Their L.C.M. is 240, which is not equal to 60.
Therefore, n = 47.
The consecutive numbers are:
47, 48, 49, 50, 51
Their L.C.M. is 240, which is not equal to 60.
Therefore, there is no solution to the problem.

Thus, the answer is None of the Above.

Answer

Let n be 1^st natural number.
So numbers be
n, n + 1, n + 2, n + 3, n + 4, n + 5
According to question
► n(n + 1) = n + 4
► n² + n = n + 4
► n² = 4
► n = 2
– 2 cannot be taken as n is natural number.
So n = 2 numbers be 2, 3, 4, 5, 6
Sum of numbers = 2 + 3 + 4 + 5 + 6 = 20.

Answer

Let n be 1^st natural number.
So numbers be
n, n + 1, n + 2, n + 3, n + 4, n + 5
According to question
► n(n + 1) = n + 4
► n² + n = n + 4
► n² = 4
► n = 2
– 2 cannot be taken as n is natural number.
So n = 2 numbers be 2, 3, 4, 5, 6
Sum of numbers = 2 + 3 + 4 + 5 + 6 = 20.

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