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# Find the unit digit of the cube root of the following number: $13824$a) $8$b) $6$c) $4$d) $2$

Last updated date: 14th Jul 2024
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We have the number $13824$. We have to find the unit digit of the cube root of this number. So, we will begin by writing the prime factorization of the number.
To write the prime factorization of any number, start by dividing the number by the first prime number, which is $2$ and then continue to divide by $2$ until you get a number which is not divisible by $2$ (which means that you get a decimal or remainder on dividing the number by $2$.
Then start dividing the number by the next prime number which is $3$. Continue dividing the number by $3$ until you get a number which is not divisible by $3$. Thus, continuing the process, keep dividing the numbers by series of prime numbers $5,7,...$ until the only numbers left are prime numbers. Write the given number as a product of all the prime numbers (considering the fact to count each prime number as many times as it divides the given number) to get the prime factorization of the given number.
Thus, we have $13824={{2}^{9}}\times {{3}^{3}}$ as the prime factorization of $13824$.
To find the cube root of $13824$, we will raise it to the power $\dfrac{1}{3}$. Thus, we have ${{\left( 13824 \right)}^{\dfrac{1}{3}}}={{\left( {{2}^{9}}\times {{3}^{3}} \right)}^{\dfrac{1}{3}}}={{\left( {{2}^{9}} \right)}^{\dfrac{1}{3}}}\times {{\left( {{3}^{3}} \right)}^{\dfrac{1}{3}}}={{2}^{\left( 9\times \dfrac{1}{3} \right)}}\times {{3}^{\left( 3\times \dfrac{1}{3} \right)}}={{2}^{3}}\times 3=8\times 3=24$ as the cube root of $13824$.
As, the cube root of $13824$ is $24$. Thus, the unit digit of the cube root of $13824$ is $4$, which is option (c).
Note: We can also find the unit digit of cube root of any number by not actually finding the cube root of the number. We just need to observe the fact that any number ending in $1,4,5,6,9$ has the unit digit of cube root as the unit digit of the number itself.