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Which of the following four digit numbers is a perfect square such that the first two digits and the last two digits considered separately also represent perfect squares ? (A) 1681 (B) 5462 (C) 8214 (D) 7210 |
Answer» Correct option is (A) 1681 (A) 1681 = 1600+80+1 \(=40^2+2\times40\times1+1^2=(40+1)^2=41^2\) \(\therefore\) 1681 is a perfect square. Also, its first two digits \(=16=4^2\) is a perfect square. And its last two digits \(=81=9^2\) is a perfect square. Thus, 1681 is a type of required number. (B) 5462 is not a perfect square because any number whose last digit (unit digit) is 2 never forms a perfect square. Thus, 5462 is not a type of required number. Similarly 8214 and 7210 are not a perfect square. Thus, both are not a type of required number. Correct option is (A) 1681 |
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