Question

What are the factors of 64?

Answer 1

Hint: We need to find the factors of 64. First, we need to find the prime factorisation 64. We take the combinations of those factorisations to form the factors of 64.

Complete step by step solution:
We first find the factors of 64.
The factors of 64 are $ 1,2,4,8,16,32,64 $ .
The process of finding the prime factorisation of a number is to find the smallest prime possible which divides the number. We complete the division then look for the same process again. We continue this process until we get 1 as the remaining quotient.
Therefore, if we need the prime factorisation of $ a $ and we find $ x $ to be the least prime that divides $ a $ , then we take $ \dfrac{a}{x} $ for the next step and find its least prime factor to continue the process.
At the end we write the prime in multiplied form to find the prime factorisation.
For our given 64, we find the prime factorisation using long division.
\[\begin{align}
  & 2\left| \!{\underline {\,
  64 \,}} \right. \\
 & 2\left| \!{\underline {\,
  32 \,}} \right. \\
 & 2\left| \!{\underline {\,
  16 \,}} \right. \\
 & 2\left| \!{\underline {\,
  8 \,}} \right. \\
 & 2\left| \!{\underline {\,
  4 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
Therefore, the prime factorisation of 64 is $ 64=2\times 2\times 2\times 2\times 2\times 2={{2}^{6}} $ .
Now we can take factors in the form of $ {{2}^{i}},i=0\left( 1 \right)6 $ .
Therefore, the factors of 64 are \[{{2}^{0}}=1,{{2}^{1}}=2,{{2}^{2}}=4,{{2}^{3}}=8,{{2}^{4}}=16,{{2}^{5}}=32,{{2}^{6}}=64\].

Note: We need to be careful about taking the factors only as primes. We can’t take composite numbers for the division. We have to find the least prime and in the ascending order to smoothly complete the factorisation.