If the integer $152AB1$ is a perfect square, what is the sum of the digits of its square root?

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If the integer $152AB1$ is a perfect square, what is the sum of the digits of its square root?

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Tài Đức 4 years 2021-07-21T16:39:26+00:00 1 Answers 96 views 0

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  1. mit
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    2021-07-21T16:41:03+00:00
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    9514 1404 393

    Answer:

      13

    Step-by-step explanation:

    152AB1 is not a square in hexadecimal, so we assume A and B are supposed to represent single digits in decimal.

    If A=B=0, √152001 ≈ 389.9

    If A=B=9, √152991 ≈ 391.1

    The least significant digit of 152AB1 being non-zero, we know it is not the square of 390. Hence, it must be the square of 391.

    For 152AB1 to be a perfect square, we must have …

      152AB1 = 391² = 152881

    The sum of the digits of the square root is 3+9+1 = 13.

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