Question

Which of the following numbers is a perfect square?

• A. 4410
• B. 2362
• C. 7921
• D. 5688
• E. 1253

Hint:

Always remember: No perfect square will ever end with the digits 2, 3, 7, or 8, nor will it ever have an odd number of trailing zeros.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to confidently identify the perfect square among the given five numbers. A perfect square is an integer that is the product of an integer multiplied by itself.


Step 2: Key Formula or Approach:
We can efficiently eliminate numbers that cannot be perfect squares by observing their unit digits. The unit digit of any perfect square can only ever be 0, 1, 4, 5, 6, or 9.
Additionally, if a perfect square ends in the digit 0, it must end in an even number of zeros (e.g., 00, 0000).


Step 3: Detailed Explanation:
Let's check the unit digit of each given option systematically:
- (B) 2362 ends in 2. It is mathematically impossible for it to be a perfect square.
- (D) 5688 ends in 8. It is mathematically impossible for it to be a perfect square.
- (E) 1253 ends in 3. It is mathematically impossible for it to be a perfect square.
We are now left with options (A) 4410 and (C) 7921.
- (A) 4410 ends in a single zero. As established, a perfect square ending in zero must have an even number of trailing zeros. Thus, 4410 is not a perfect square (though 441 is \( 21^2 \), 4410 is \( 21^2 \times 10 \)).
- By process of elimination, (C) 7921 must be the correct perfect square.
To verify this conclusively, let's estimate its square root. We know that \( 80^2 = 6400 \) and \( 90^2 = 8100 \). The number 7921 is very close to 8100 and ends in 1, meaning its square root must end in 1 or 9. Let's test 89:
\[ 89^2 = (90 - 1)^2 = 90^2 - 2(90)(1) + 1^2 = 8100 - 180 + 1 = 7921 \]
Thus, 7921 is indeed the perfect square of 89.


Step 4: Final Answer:
The perfect square is 7921.