Find the smallest number which when added by 6 is exactly divisible by 9 and 33.

Explanation

To solve the problem of finding the smallest number which, when added to 6, is exactly divisible by both 9 and 33, we can follow these steps:

### Step 1: Find the Least Common Multiple (LCM) of 9 and 33
To determine the smallest number that is divisible by both 9 and 33, we first need to find their LCM.

- The prime factorization of 9 is \(3^2\).
- The prime factorization of 33 is \(3^1 \times 11^1\).

To find the LCM, we take the highest power of each prime factor:
- For 3, the highest power is \(3^2\).
- For 11, the highest power is \(11^1\).

Thus, the LCM is:
\[ LCM = 3^2 \times 11^1 = 9 \times 11 = 99 \]

### Step 2: Set Up the Equation
Let \(x\) be the smallest number we are looking for. According to the problem, when we add 6 to this number, it should be divisible by both 9 and 33. Therefore, we can set up the equation:
\[ x + 6 = 99k \]
where \(k\) is any integer.

### Step 3: Solve for \(x\)
To find \(x\), we rearrange the equation:
\[ x = 99k - 6 \]
To find the smallest positive \(x\), we can start with \(k = 1\):
\[ x = 99(1) - 6 = 99 - 6 = 93 \]

### Step 4: Verify the Solution
Now, we need to check if \(93 + 6\) is divisible by both 9 and 33:
\[ 93 + 6 = 99 \]
Now, check divisibility:
- \(99 \div 9 = 11\) (which is an integer)
- \(99 \div 33 = 3\) (which is also an integer)

Since both conditions are satisfied, our solution is correct.

### Final Answer
The smallest number is **93**.

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