Question

The remainder of any perfect square divided by 3 is
(a) 0
(b) 1
(c) Either (a) or (b)
(d) Neither (a) nor (b)

Answer 1

Hint: We divide the numbers into three types on the basis of their divisibility by 3. Use the concept of perfect squares to calculate the perfect squares of the three numbers. Take the square of each number and divide by 3 to find the remainder in each case.
* A number is a perfect square if on taking the square root of the perfect square we get a natural number.
* If ‘a’ is divided by ‘b’ and the quotient comes out to be ‘q’ and the remainder is ‘r’, then we can write \[a = bq + r\]

Complete step-by-step answer:
We have to find the remainder of a perfect square when it is divided by 3
So, we divide the set of natural numbers in three types of numbers i.e.\[3n,3n + 1,3n + 2\]
We know a number is the perfect square of a number if it is a square of a natural number.
So, we calculate perfect squares of each of three numbers
\[ \Rightarrow {\left( {3n} \right)^2}\] is a perfect square of the number \[3n\]
\[{\left( {3n + 1} \right)^2}\]is a perfect square of the number \[3n + 1\]
\[{\left( {3n + 2} \right)^2}\]is a perfect square of the number \[3n + 2\]
We solve for the remainder of each of the perfect squares when divided by 3 separately.
CASE 1:
We have to divide the number \[{\left( {3n} \right)^2}\]by 3
\[ \Rightarrow {\left( {3n} \right)^2} = 9{n^2}\]
Write the RHS of the equation as a factor of 3.
\[ \Rightarrow {\left( {3n} \right)^2} = 3 \times \left( {3{n^2}} \right)\]
Compare the equation with the general equation of division i.e. \[a = bq + r\]
\[ \Rightarrow {\left( {3n} \right)^2} = 3 \times \left( {3{n^2}} \right) + 0\]
Remainder \[ = 0\]
CASE 2:
We have to divide the number \[{\left( {3n + 1} \right)^2}\]by 3
Use the formula \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]to open the square term.
\[ \Rightarrow {\left( {3n + 1} \right)^2} = {(3n)^2} + {(1)^2} + 2 \times (3n) \times (1)\]
\[ \Rightarrow {\left( {3n + 1} \right)^2} = 9{n^2} + 1 + 6n\]
Take common 3 from the possible terms and pair them
\[ \Rightarrow {\left( {3n + 1} \right)^2} = 3(3{n^2} + 2n) + 1\]
Compare the equation with the general equation of division i.e. \[a = bq + r\]
Remainder \[ = 1\]
CASE 3:
We have to divide the number \[{\left( {3n + 2} \right)^2}\]by 3
Use the formula \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]to open the square term.
\[ \Rightarrow {\left( {3n + 2} \right)^2} = {(3n)^2} + {(2)^2} + 2 \times (3n) \times (2)\]
\[ \Rightarrow {\left( {3n + 2} \right)^2} = 9{n^2} + 4 + 12n\]
Break the constant value in such a way that one part of it is a factor of 3
\[ \Rightarrow {\left( {3n + 2} \right)^2} = 9{n^2} + 12n + 3 + 1\]
Take common 3 from the possible terms and pair them
\[ \Rightarrow {\left( {3n + 2} \right)^2} = 3(3{n^2} + 4n + 1) + 1\]
Compare the equation with the general equation of division i.e. \[a = bq + r\]
Remainder \[ = 1\]
From the above three cases, the remainder is either 0 or 1.
\[\therefore \]Remainder of any perfect square when divided by 3 is either 0 or 1

\[\therefore \]Option C is correct.

Note: Students might try to solve this question by taking square of particular numbers starting from1 and then dividing each perfect square by 3, but this process is not helpful here because we have to find the remainder for all perfect squares and it will take a long time to calculate for each number.
Keep in mind we use particular numbers mostly when we have to prove a contradiction to any statement. Also, in general we can divide the set of natural numbers to any type of numbers like \[an,an + b,an + c,...\] until the point where the second number in addition becomes a multiple of ‘a’.