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

9514 1404 393

  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.

y > 1/3x

y < = -x + 4

hope this : )

5/12, or 5: 12

step-by-step explanation: