IF n is a three digit number and last two digits of square of n are `54(n^(2) = …. 54),` then how many values of n are possible ?

Options

A. 2
B. 20
C. 1
D. 0

Answer

To solve the problem, we need to find all three-digit numbers \( n \) such that the last two digits of \( n^2 \) are 54. This means we need to find \( n \) such that:

\[ n^2 \equiv 54 \mod{100} \]

### Step 1: Set up the congruence
We start by setting up the congruence:

\[ n^2 \equiv 54 \mod{100} \]

### Step 2: Find possible values of \( n \)
To find the values of \( n \), we can check the squares of numbers ending with digits from 0 to 9, as these will determine the last two digits of \( n^2 \).

- \( 0^2 = 0 \)
- \( 1^2 = 1 \)
- \( 2^2 = 4 \)
- \( 3^2 = 9 \)
- \( 4^2 = 16 \)
- \( 5^2 = 25 \)
- \( 6^2 = 36 \)
- \( 7^2 = 49 \)
- \( 8^2 = 64 \)
- \( 9^2 = 81 \)

Now, we will check which of these squares can give us a last two digits of 54.

### Step 3: Check squares of numbers ending with specific digits
We can check the last digits of \( n \) to see if any of them can yield \( n^2 \equiv 54 \mod{100} \):

- \( n \equiv 2 \mod{10} \) gives \( n^2 \equiv 4 \)
- \( n \equiv 4 \mod{10} \) gives \( n^2 \equiv 16 \)
- \( n \equiv 6 \mod{10} \) gives \( n^2 \equiv 36 \)
- \( n \equiv 8 \mod{10} \) gives \( n^2 \equiv 64 \)

None of these yield \( n^2 \equiv 54 \mod{100} \).

### Step 4: Conclude the results
Since none of the possible last digits of \( n \) yield a square that ends with 54, we conclude that there are no three-digit numbers \( n \) such that \( n^2 \) ends with 54.

Thus, the number of values of \( n \) is:

\[ \text{Number of values of } n = 0 \]

### Final Answer
The answer is \( 0 \).

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