Explanation
Given, the number is $$143^3$$.
Here, the units digit is $$3$$.
We know, the cube of $$3$$, i.e. $$3^3=27$$, whose units place is $$7$$.
Therefore, the units digit of the cube of $$143$$, i.e. $$143^3$$ is $$7$$.
Hence, option $$C$$ is correct.
Given, the number is $$98765^3$$.
Here, the units digit is $$5$$.
We know, the cube of $$5$$, i.e. $$5^3=125$$, whose units place is $$5$$.
Therefore, the units digit of the cube of $$98765$$, i.e. $$98765^3$$ is $$5$$.
Hence, option $$A$$ is correct.
Given, the number is $$128^3$$.
Here, the units digit is $$8$$.
We know, the cube of $$8$$, i.e. $$8^3=512$$, whose units place is $$2$$.
Therefore, the units digit of the cube of $$128$$, i.e. $$128^3$$ is $$2$$.
Hence, option $$D$$ is correct.
Given, the units digit of cube of a number is $$6$$.Any number having $$6$$ in it's unit's place will have a $$6$$ in the unit's place of cube.It can be observed $$...6\times ...6\times ...6=......6$$
Therefore if the unit's digit of the cube is $$6$$, the number must have $$6$$ in it's unit's place.
Given, the number is $$100001^3$$.
Here, the units digit is $$1$$.
We know, the cube of $$1$$, i.e. $$1^3=1$$, whose units place is $$1$$.
Therefore, the units digit of the cube of $$100001$$, i.e. $$100001^3$$ is $$1$$.
We know,
$$\displaystyle 1=1={ 1 }^{ 3 }$$,
$$\displaystyle 3+5=8={ 2 }^{ 3 }$$,
$$\displaystyle 7+9+11=27={ 3 }^{ 3 }$$.
On prime factorisation of the numbers individually, we get,
$$512=\underline { 2 \times 2 \times 2 } \times \underline { 2 \times 2 \times 2 } \times \underline { 2 \times 2 \times 2 }$$
$$=2^3 \times 2^3 \times 2^3=8^3$$.
$$125=\underline { 5 \times 5 \times 5 }$$ $$ =5^3$$.
Therefore,
$$\sqrt[3]{4\dfrac{12}{125}}=\sqrt[3]{\dfrac{512}{125}}$$
$$=\sqrt[3]{\dfrac{8^3}{5^3}}={\dfrac{8}{5}}$$ $$={1\dfrac{3}{5}}$$.
Hence, option $$B$$ is correct.
$$108=3\times 3\times 3\times 2\times 2$$
$$= 3 ^3 \times 2 ^2$$.
We know, a perfect cube has multiples of $$3$$ as powers of prime factors.
Here, number of $$3$$'s is $$3$$ and number of $$2$$'s is $$2$$.
So we need to multiply another $$2$$ to the factorization to make $$108$$ a perfect cube.
Hence, the smallest number by which $$108$$ must be multiplied to obtain a perfect cube is $$2$$.
$$8=\underline { 2 \times 2 \times 2 }$$
$$125=\underline { 5 \times 5 \times 5 }$$
$$ \Rightarrow \sqrt [ 3 ]{ 125 } =5$$.
$$64=\underline { 2 \times 2 \times 2 } \times \underline { 2 \times 2 \times 2 }$$
$$ \Rightarrow \sqrt [ 3 ]{ 64 } =2 \times 2=4$$.
Prime factorising $$1440$$, we get,
$$ 1440 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5$$
$$= 2 ^5 \times 3 ^2 \times 5^1$$.
Here, number of $$2$$'s is $$5$$, number of $$3$$'s is $$2$$ and number of $$5$$'s is $$1$$.
So we need to multiply another $$2$$, $$3$$ and $$5^2$$ in the factorization to make $$1440$$ a perfect cube.
Hence, the smallest number by which $$1440$$ must be multiplied to obtain a perfect cube is $$2\times 3 \times 5^2=150$$.
$$\therefore$$ The sum of digits of the smallest number is $$= 1+5+0 = 6$$.
Prime factorising $$9000$$, we get,
$$ 9000 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 5$$
$$= 2^3 \times 3 ^2 \times 5^3$$
We know, a perfect cube has prime factors in powers of $$3$$.
Here, number of $$2$$'s is $$3$$, number of $$3$$'s is $$2$$ and number of $$5$$'s is $$3$$.
So we need to remove $$3^2$$ from the factorization to make $$9000$$ a perfect cube.
Hence, the smallest number by which $$9000$$ must be divided to obtain a perfect cube is $$3^2=9$$.
Please disable the adBlock and continue. Thank you.
