How many natural numbers upto 990 are divisible by 5 & 9 both, but not by 7?

Options

A. A)18
B. B)19
C. C)22
D. D)none of a,b,c

Answer

To solve the problem of finding how many natural numbers up to 990 are divisible by both 5 and 9 but not by 7, we can follow these steps:

### Step 1: Find the Least Common Multiple (LCM) of 5 and 9
To find the numbers that are divisible by both 5 and 9, we first need to find their LCM.

- The prime factorization of 5 is \(5^1\).
- The prime factorization of 9 is \(3^2\).

The LCM is calculated by taking the highest power of each prime factor:
\[ \text{LCM}(5, 9) = 5^1 \times 3^2 = 5 \times 9 = 45 \]

### Step 2: Count the Multiples of 45 up to 990
Next, we need to find how many multiples of 45 are there up to 990. This can be calculated by dividing 990 by 45:
\[ \text{Number of multiples of 45} = \left\lfloor \frac{990}{45} \right\rfloor \]
Calculating this gives:
\[ \frac{990}{45} = 22 \]
So, there are 22 multiples of 45 up to 990.

### Step 3: Find the Multiples of 315 (LCM of 45 and 7)
Now, we need to exclude those multiples of 45 that are also divisible by 7. To do this, we find the LCM of 45 and 7:
\[ \text{LCM}(45, 7) = 45 \times 7 = 315 \]

Now, we count how many multiples of 315 are there up to 990:
\[ \text{Number of multiples of 315} = \left\lfloor \frac{990}{315} \right\rfloor \]
Calculating this gives:
\[ \frac{990}{315} \approx 3.14 \Rightarrow \left\lfloor 3.14 \right\rfloor = 3 \]
So, there are 3 multiples of 315 up to 990.

### Step 4: Subtract the Count of Multiples of 315 from Multiples of 45
Finally, we subtract the count of multiples of 315 from the count of multiples of 45 to find the numbers that are divisible by 5 and 9 but not by 7:
\[ \text{Natural numbers divisible by 5 and 9 but not by 7} = 22 - 3 = 19 \]

### Conclusion
Thus, the total number of natural numbers up to 990 that are divisible by both 5 and 9 but not by 7 is **19**.

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