Q: By what smallest natural number should 32 be multiplied to get a perfect square?

Prime factorisation of 32: $$32 = 2 \times 2 \times 2 \times 2 \times 2 = (2 \times 2) \times (2 \times 2) \times 2$$ \nHere, one $2$ is unpaired. So, 32 must be multiplied by 2 to form pairs.

Question 1

What is the unit digit of the square of the number 67?

Accepted Answer

To find the unit digit of $67^2$, we look at the unit digit of $67$, which is $7$. 
Since $7^2 = 49$, the unit digit is 9.

Question 2

How many non-square numbers lie between $12^2$ and $13^2$?

Accepted Answer

The number of non-square numbers between $n^2$ and $(n+1)^2$ is given by the formula $2n$. 
Here $n = 12$, so: $$	ext{Number of non-square numbers} = 2 	imes 12 = 24$$

Question 3

Which of the following is a Pythagorean triplet whose smallest member is 6?

Accepted Answer

A Pythagorean triplet is of the form $(2m, m^2 - 1, m^2 + 1)$. 
Given smallest member $2m = 6 \implies m = 3$. Second member: $m^2 - 1 = 3^2 - 1 = 8$ Third member: $m^2 + 1 = 3^2 + 1 = 10$ 
Thus, the triplet is (6, 8, 10).

Question 4

By what smallest natural number should 32 be multiplied to get a perfect square?

Accepted Answer

Prime factorisation of 32: $$32 = 2 	imes 2 	imes 2 	imes 2 	imes 2 = (2 	imes 2) 	imes (2 	imes 2) 	imes 2$$ 
Here, one $2$ is unpaired. So, 32 must be multiplied by 2 to form pairs.

Question 5

Which of the following digits can NEVER be at the unit's place of a perfect square number?

Accepted Answer

A perfect square number always ends in $0, 1, 4, 5, 6,$ or $9$. 
It NEVER ends in 2, 3, 7, or 8.
Therefore, 7 cannot be at the unit's place of a perfect square.

Square and Square Roots