# Square root of perfect squares

This question was given In the intermediate category of the 2^{nd} Philippine National Vedic Math Olympiad last 2021 were participants are required to answer 50 questions in 1 hour.

Since all the choices are whole numbers, we can assume that 7056 is a perfect square.

If we know our table of squares well, we can easily solve this problem using the Vedic Math sutra **the First by the First and the Last by the Last**.

Table of squares

1^{2} = 1 9^{2} = 81

2^{2} = 4 8^{2} = 64

3^{2} = 9 7^{2} = 49

4^{2} = 16 6^{2} = 36

5^{2} = 25

From the table of squares, we can see that the squares of ten’s complement like 1 and 9, 2 and 8, etc. end in the same digits.

We can split 7056 into two groups of two digits each. Since the first group, 70 is more than 64 (square of 8) but less than 81 (the square of 9), the first digit must be 8.

But from the last digit of the second group, 56, we can see that we have two possible answers: 84 and 86. How do we quickly determine which is the correct one?

From our post last week, we learned that squares of numbers ending in 5 follow the **By One more than the One Before** sutra. The first part of the square of 85 is obtained by multiplying 8 by **one more than 8**, that is, 8 x 9 = 72. Since 70 is less than the 72, the square root of 7056 is 84.

Exercises: Find the square root of the following perfect squares:

- 576 →
- 1024 →
- 1521 →
- 1849 →
- 2916 →
- 4356 →
- 5041 →
- 6084 →
- 7744 →
- 8836 →

Answers to last week’s exercises:

Find the square root of the following numbers ending in 25

- 625 → 25
- 7225 → 85
- 9025 → 95
- 225 → 15
- 3025 → 55
- 2025 → 45
- 11025 → 105
- 15625 → 125
- 13225 → 115
- 990,025 → 995