[x/2]+[x/3]+[x/5]=x where [.] denotes gif .Find no of real solutions.

To solve the equation \([x/2] + [x/3] + [x/5] = x\), where \([.]\) denotes the greatest integer function (also known as the floor function), we need to analyze how the floor function behaves with respect to the variable \(x\). The floor function \([x]\) gives the largest integer less than or equal to \(x\). This means that for any real number \(x\), \([x/n]\) will yield an integer value that is less than or equal to \(x/n\).

Breaking Down the Equation

Let's rewrite the equation in a more manageable form. We can express it as:

\[ [x/2] + [x/3] + [x/5] = x \]

Since \([x/n]\) is always less than or equal to \(x/n\), we can derive some inequalities:

• \([x/2] \leq x/2\)
• \([x/3] \leq x/3\)
• \([x/5] \leq x/5\)

Adding these inequalities gives:

\[ [x/2] + [x/3] + [x/5] \leq x/2 + x/3 + x/5 \]

Finding a Common Denominator

The right-hand side can be simplified by finding a common denominator, which in this case is 30:

\[ \frac{15x}{30} + \frac{10x}{30} + \frac{6x}{30} = \frac{31x}{30} \]

This leads us to:

\[ [x/2] + [x/3] + [x/5] \leq \frac{31x}{30} \]

Establishing the Range for \(x\)

For the equation to hold true, we also need to consider the fact that \([x/2] + [x/3] + [x/5]\) must be an integer, while \(x\) is a real number. This means that \(x\) must be such that the left-hand side can equal \(x\). Since \([x/2] + [x/3] + [x/5]\) is always less than or equal to \(\frac{31x}{30}\), we can set up the following inequality:

\[ x \leq \frac{31x}{30} \]

Rearranging gives:

\[ 30x \leq 31x \Rightarrow 0 \leq x \]

Finding Integer Solutions

Next, we need to find specific integer values of \(x\) that satisfy the original equation. Since \([x/2]\), \([x/3]\), and \([x/5]\) are all integers, we can test integer values for \(x\). Let’s denote \(x\) as \(n\), where \(n\) is a non-negative integer:

• For \(n = 0\): \([0/2] + [0/3] + [0/5] = 0\) (solution)
• For \(n = 1\): \([1/2] + [1/3] + [1/5] = 0\) (not a solution)
• For \(n = 2\): \([2/2] + [2/3] + [2/5] = 1 + 0 + 0 = 1\) (not a solution)
• For \(n = 3\): \([3/2] + [3/3] + [3/5] = 1 + 1 + 0 = 2\) (not a solution)
• For \(n = 4\): \([4/2] + [4/3] + [4/5] = 2 + 1 + 0 = 3\) (not a solution)
• For \(n = 5\): \([5/2] + [5/3] + [5/5] = 2 + 1 + 1 = 4\) (not a solution)
• For \(n = 6\): \([6/2] + [6/3] + [6/5] = 3 + 2 + 1 = 6\) (solution)
• For \(n = 7\): \([7/2] + [7/3] + [7/5] = 3 + 2 + 1 = 6\) (not a solution)
• For \(n = 8\): \([8/2] + [8/3] + [8/5] = 4 + 2 + 1 = 7\) (not a solution)
• For \(n = 9\): \([9/2] + [9/3] + [9/5] = 4 + 3 + 1 = 8\) (not a solution)
• For \(n = 10\): \([10/2] + [10/3] + [10/5] = 5 + 3 + 2 = 10\) (solution)

Identifying the Solutions

From our testing, we found that the integer solutions are \(n = 0\), \(n = 6\), and \(n = 10\). Thus, the total number of real solutions to the equation \([x/2] + [x/3] + [x/5] = x\) is:

Three real solutions: \(0\), \(6\), and \(10\).