Question

Using the prime factorization method, find which of the following numbers are not perfect squares.
A. 400
B. 768
C. 1296
D. 8000
E. 9025

Answer 1

Hint: Here in this question we need to find whether the given numbers are perfect squares using prime factorization. By expressing the given number as a product of its prime factors and then group all of the factors in pairs such that both the factors in each pair are equal. This pairing leads us to the perfect square, whose exact square root can be obtained.

Complete step by step answer:
To know which numbers are not perfect squares, firstly we need to find the prime factors of each number. Then pair them with the same facto, in this process no factor should be left without a pair.

(A) Considering the first number 400, the prime factors of this number are
\[ 400 = 2 \times \]2\[ \times \]2\[ \times \]2\[ \times \]5\[ \times \]5
On pairing the prime factors
\[400 = \underline {2 \times 2} \times \underline {2 \times 2} \times \underline {5 \times 5} \]
\[ \Rightarrow 400 = \]2 \[ \times \]2 \[ \times \]5
\[ \Rightarrow 400 = \]20
The prime factors are all paired, thus we can say that 400 is a perfect square.

(B) Considering the second number 768, the prime factors of this number are
\[ \Rightarrow 768 = \]2 \[ \times \]2 \[ \times \]2 \[ \times \]2 \[ \times \]2 \[ \times \]2 \[ \times \]2 \[ \times \]2 \[ \times \]3
On pairing the prime factors
\[ \Rightarrow 768 = \] \[\underline {2 \times 2} \times \underline {2 \times 2} \times \underline {2 \times 2} \times \underline {2 \times 2} \times 3\]
Here the prime factor 3 is left without pairing, thus 768 is not a perfect square.

(C) Considering the second number 1296, the prime factors of this number are
\[ \Rightarrow 1296 = \]2 \[ \times \]2 \[ \times \]2 \[ \times \]2 \[ \times \]3\[ \times \]3\[ \times \] 3\[ \times \]3
On pairing the prime factors
\[ \Rightarrow 1296 = \] \[\underline {2 \times 2} \times \underline {2 \times 2} \times \]\[\underline {3 \times 3} \]\[ \times \]\[\underline {3 \times 3} \]
\[ \Rightarrow 1296 = \]2\[ \times \]2 \[ \times \]3\[ \times \]3
\[ \Rightarrow 1296 = \]36
The prime factors are all paired, thus we can say that 1296 is a perfect square.

(D) Considering the second number 8000, the prime factors of this number are
\[ \Rightarrow 8000 = \] 2\[ \times \]2 \[ \times \]2 \[ \times \]2 \[ \times \]2 \[ \times \]2 \[ \times \]5 \[ \times \]5 \[ \times \]5
On pairing the prime factors
\[ \Rightarrow 8000 = \] \[\underline {2 \times 2} \times \underline {2 \times 2} \times \underline {2 \times 2} \times \]\[\underline {5 \times 5} \]\[ \times 5\]
Here the prime factor 5 is left without pairing, thus 8000 is not a perfect square.

(E) Considering the second number 9025, the prime factors of this number are
\[ \Rightarrow 9025 = \]5\[ \times \]5\[ \times \]19\[ \times \]19
On pairing the prime factors
\[ \Rightarrow 9025 = \]\[\underline {5 \times 5} \times \underline {19 \times 19} \]
\[ \Rightarrow 9025 = \]5 \[ \times \]19
\[ \Rightarrow 9025 = \]95
The prime factors are all paired, thus we can say that 9025 is a perfect square.

Thus 768 and 8000 are not the perfect squares.

Note:The given question can be solved by writing the prime factors in the index form. Then we take half of each index value in order to find the required square root.Similarly, we can use the same method to find the perfect cube. Remember the square root of a perfect square of an even number is always an even number and the square root of a perfect square of an odd number is always an odd number.