If a perfect square of a number contains 22 digits, then how many digits will it have ?

Options

A. 12
B. 11
C. 9
D. 10

Answer

To determine how many digits a number has if its perfect square contains 22 digits, we can follow these steps:

### Step 1: Understand the relationship between the number of digits and the perfect square.
The number of digits \(d\) in a number \(x\) can be calculated using the formula:
\[ d = \lfloor \log_{10} x \rfloor + 1 \]
For a perfect square \(y = x^2\), if \(y\) has \(n\) digits, we can express this as:
\[ n = \lfloor \log_{10} (x^2) \rfloor + 1 \]

### Step 2: Relate the digits of the perfect square to the original number.
Using the properties of logarithms, we can rewrite the equation for the perfect square:
\[ n = \lfloor 2 \log_{10} x \rfloor + 1 \]
If \(y\) has \(n\) digits, then:
\[ n - 1 = \lfloor 2 \log_{10} x \rfloor \]

### Step 3: Substitute the given number of digits.
Given that the perfect square has 22 digits, we can substitute \(n = 22\):
\[ 22 - 1 = \lfloor 2 \log_{10} x \rfloor \]
This simplifies to:
\[ 21 = \lfloor 2 \log_{10} x \rfloor \]

### Step 4: Solve for \( \log_{10} x \).
From the equation \(21 = \lfloor 2 \log_{10} x \rfloor\), we can deduce that:
\[ 21 \leq 2 \log_{10} x < 22 \]
Dividing the entire inequality by 2 gives:
\[ 10.5 \leq \log_{10} x < 11 \]

### Step 5: Convert back to the number of digits in \(x\).
Now we can find the number of digits in \(x\):
\[ \lfloor \log_{10} x \rfloor + 1 \]
Since \(10.5 \leq \log_{10} x < 11\), we have:
\[ \lfloor \log_{10} x \rfloor = 10 \]
Thus, the number of digits in \(x\) is:
\[ 10 + 1 = 11 \]

### Conclusion:
Therefore, if a perfect square of a number contains 22 digits, the original number will have **11 digits**.

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