Finding the Square root through Prime Factorisation Method

Finding square root through prime factorisation

Consider the prime factorisation of the following numbers and their squares:

How many times does 2 occur in the prime factorisation of 6? .

How many times does 2 occur in the prime factorisation of 36? .

Similarly, observe the occurrence of 3 in 6 and 36, of 2 in 8 and 64 etc.

You will find that each prime factor in the prime factorisation of the square of a number, occurs twice the number of times it occurs in the prime factorisation of the number itself.

Let us use this to find the square root of a given square number, say 324.

We know that the prime factorisation of 324 is

324 = × × × × ×

By pairing the prime factors, we get

324 = 2 × 2 × 3 × 3 × 3 × 3 = 22 × 32 × 32 = 2×3×32

So,

324 = 2 × 3 × 3 =

Similarly can you find the square root of 256? Prime factorisation of 256 is:

256 = × × × × × × ×

By pairing the prime factors we get,

256 = (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) = 2×2×2×22

Therefore,

256 = 2 × 2 × 2 × 2 =

Is 48 a perfect square?

We know,

48 = (2 × 2) × (2 × 2) × 3

Since all the factors are not in pairs so 48 is not a perfect square.

Suppose we want to find the smallest multiple of 48 that is a perfect square, how should we proceed?

Making pairs of the prime factors of 48 we see that is the only factor that does not have a pair.

So we need to multiply by 3 to complete the pair.

Hence:

48 × 3 = which is a perfect square.

Can you tell by which number should we divide 48 to get a perfect square?

The factor is not in a pair in the factorisation of 48.

So, if we divide 48 by 3 we get 48 ÷ 3 =

and this number 16 is a perfect square too.

Example 4: Find the square root of 6400.

Writing prime factorisation for 6400: 6400 = × × × × × × × × ×

Therefore, 6400 = × × × × =

Example 5: Is 90 a perfect square?

We have: 90 = × × ×

The prime factors 2 and 5 do not occur in pairs.

Therefore, 90 is not a square.

That 90 is not a perfect square can also be seen from the fact that it has only one zero.

Example 6: Is 2352 a perfect square? If not, find the smallest multiple of 2352 which is a perfect square. Find the square root of the new number.

We have: 2352 = × × × × × ×

As the prime factor 3 has no pair, 2352 is not a perfect square.

If 3 gets a pair then the number will become perfect square. So, we multiply 2352 by 3 to get,

2352 × 3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7

Now each prime factor is in a pair.

Therefore, 2352 × 3 = is a perfect square.

Thus, the required smallest multiple of 2352 is 7056 which is a perfect square.

7056 = × × × =

Example 7: Find the smallest number by which 9408 must be divided so that the quotient is a perfect square. Find the square root of the quotient.

We have, 9408 = × × × × × × × ×

If we divide 9408 by the factor 3, then

9408 ÷ 3 = = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7

which is a perfect square.

Therefore, the required smallest number by which 9408 needs to be divided, to become a perfect square, is .

And,

3136 = × × × =

Example 8: Find the smallest square number which is divisible by each of the numbers 6, 9 and 15.

This has to be done in two steps. First, find the smallest common multiple and then find the square number needed. The least number divisible by each one of 6, 9 and 15 will be their LCMHCF.

The LCM of 6, 9 and 15 is = × × ×

We see that prime factors 2 and 5 are not in pairs.

Therefore, 90 is not a perfect square.

In order to get a perfect square, each factor of 90 must be paired. So we need to make pairs of and .

Therefore, 90 should be multiplied by .

Hence, the required square number is 90 × 10 = .

Prime Factorization Finder