Question

If 0 < x < 1000 and \[\left[ \dfrac{x}{2} \right]+\left[ \dfrac{x}{3} \right]+\left[ \dfrac{x}{5} \right]=\dfrac{31}{30}x\] where \[\left[ x \right]\] is greatest integer function, then the number of possible values of x is

(a) 34

(b) 33

(c) 32

(d) None of these

Answer 1

Hint: First convert the integer part into the fractional part. Find the condition on the fractional part. By this find the condition on x. Apply this condition to the given range of x. From this, you can get all the possible values of x which is the required result.

Complete step-by-step answer:

The given condition in the question is written as:

\[\left[ \dfrac{x}{2} \right]+\left[ \dfrac{x}{3} \right]+\left[ \dfrac{x}{5} \right]=\dfrac{31}{30}x\]

The relation between the greatest integer function and fractional part is given by:

\[\left[ x \right]+\left\{ x \right\}=x\]

By subtraction {x} on both the sides, we get the equation as

\[\left[ x \right]+\left\{ x \right\}-\left\{ x \right\}=x-\left\{ x \right\}\]

By simplifying the above equation, we get it as

\[\left[ x \right]=x-\left\{ x \right\}\]

By substituting this relation into our given expression, we get,

\[\dfrac{x}{2}-\left\{ \dfrac{x}{2} \right\}+\dfrac{x}{3}-\left\{ \dfrac{x}{3} \right\}+\dfrac{x}{5}-\left\{ \dfrac{x}{5} \right\}=\dfrac{31x}{30}\]

Now, combining the similar terms together, we get the equation as

\[\left( \dfrac{x}{2}+\dfrac{x}{3}+\dfrac{x}{5} \right)-\left( \left\{ \dfrac{x}{2} \right\}+\left\{ \dfrac{x}{3} \right\}+\left\{ \dfrac{x}{5} \right\} \right)=\dfrac{31x}{30}\]

By substituting the value of the first term, we can write the equation as

\[\dfrac{31x}{30}-\left( \left\{ \dfrac{x}{2} \right\}+\left\{ \dfrac{x}{3} \right\}+\left\{ \dfrac{x}{5} \right\} \right)=\dfrac{31x}{30}\]

By subtracting \[\dfrac{31x}{30}\] on both sides, we get the equation as

\[\dfrac{31x}{30}-\dfrac{31x}{30}-\left( \left\{ \dfrac{x}{2} \right\}+\left\{ \dfrac{x}{3} \right\}+\left\{ \dfrac{x}{5} \right\} \right)=\dfrac{31x}{30}-\dfrac{31x}{30}\]

By simplifying the above equation, we get,

\[\left\{ \dfrac{x}{2} \right\}+\left\{ \dfrac{x}{3} \right\}+\left\{ \dfrac{x}{5} \right\}=0\]

From this equation, we can say that x must satisfy

\[\left\{ \dfrac{x}{2} \right\}=0;\left\{ \dfrac{x}{3} \right\}=0;\left\{ \dfrac{x}{5} \right\}=0\]

From these conditions, we can say the statement of x as x is a number which is a multiple of 2, 3, and 5. Now we need the least common multiple of 2, 3 and 5.

Prime factorization of 2: \[2=2\times 1\]

Prime factorization of 3: \[3=3\times 1\]

Prime factorization of 5: \[5=5\times 1\]

As all of the three are prime numbers, we can say

\[LCM\left( 2,3,5 \right)=2\times 3\times 5\]

By simplifying the above equation, we get,

\[LCM\left( 2,3,5 \right)=30\]

As x is a multiple of these 3, we can say that x = 30m. We are given that x is in the range of 0 < x < 1000. By substituting our value into the range, we get,

0 < 30m < 1000

By dividing 30 on the whole inequality, we get,

0 < m < 33.33

By this, we get the value of m in the range as

\[m\in \left[ 1,33 \right]\]

So, the number of integers in this range is 33, as there are 33 different possibilities for m. Then x will also have 33 different possibilities.

Hence, option (b) is the right answer.

Note: While changing into the fractional part, we must take care of the ‘–‘ sign. The fractional part implies that the number is multiple. So, the number to be a multiple of 3, the number must be a multiple of the LCM. This idea is very important as it is the base of our solution. So, do these steps carefully