Squares and Square Roots Test - 2

This is the correct answer.

For any natural number m, we know that 2 m, m² - 1, m² + 1 is a Pythagorean triplet.

Here, 2 m = 14
or, m = 7

∴ m² - 1
= 49 - 1
= 48

and, m² + 1
= 49 + 1
= 50

Thus, 14, 48 and 50 are Pythagorean triplets.

This is the correct answer.
For any natural number m>1, we know that 2 m, m² - 1 and m² + 1 is a

Pythagorean triplet.
Here, 2 m = 16
or, m = 8

∴ m² - 1
= 64 - 1
= 63

and, m² + 1
= 64 + 1
= 65

Thus, 63 and 65 are the Pythagorean triplets of 16.

This is the correct answer.

Resolving 127008 into prime factors, we get
127008 = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 7 x 7

Grouping the factors into pairs of equal factors, we get
127008 = (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x (7 x 7) x 2

For a number to be a perfect square, it should be possible to pair all its prime factors. In this case, 2 is without a pair.
Thus, the number must be multiplied by 2 so that the product is a perfect square.

Hence, the smallest number by which 127008 should be multiplied is 2.

This is the correct answer.

The smallest number which is divisible by each of the numbers 7, 8 and 9 is their LCM.

The LCM of 7, 8 and 9 is 4 x 2 x 9 x 7 = 504

Resolving 504 into prime factors, we get
504 = 2 x 2 x 2 x 3 x 3 x 7

Grouping the factors into pairs of equal factors, we get
504 = (2 x 2) x 2 x (3 x 3) x 7

Hence, there is no factor to form a pair with 2 and 7.
Thus, to make 504 a perfect square, it must be multiplied it by 2 x 7 = 14.

Hence, the smallest square number which is divisible by each of the numbers 7, 8 and 9 is 504 x 14 = 7056.

Sum of first n odd numbers = n²
∴ Sum of first 75 odd numbers = (75)² = 5625

2423 x 2425 = a² - 1 = (a - 1) (a + 1)
(2424 - 1)(2424 + 1) = (a - 1)(a + 1)

This gives a = 2424

9² = 81

Since 9² ends with the digit 1, so (73219)² will also end with the digit 1. None of the other options ends up as a number whose unit's digit is 1.