The number of all 2-digit numbers n such that | vedclass

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(C) Let the $2$-digit number be $n = 10a + b$,where $a \in \{1, 2, \dots, 9\}$ and $b \in \{0, 1, \dots, 9\}$.Given,$n = a^2 + b^3$.Therefore,$10a + b

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$2$

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(C) Let the $2$-digit number be $n = 10a + b$,where $a \in \{1, 2, \dots, 9\}$ and $b \in \{0, 1, \dots, 9\}$.Given,$n = a^2 + b^3$.Therefore,$10a + b = a^2 + b^3$.Rearranging the terms,we get $a^2 - 10a + b^3 - b = 0$,which can be written as $a(10 - a) = b^3 - b = b(b - 1)(b + 1)$.We test values for $b$:If $b = 0$,$a(10 - a) = 0 \Rightarrow a = 0$ (not a $2$-digit number) or $a = 10$ (not a digit).If $b = 1$,$a(10 - a) = 0$,which gives no valid $a$.If $b = 2$,$a(10 - a) = 2(1)(3) = 6$,which gives no integer solution for $a$.If $b = 3$,$a(10 - a) = 3(2)(4) = 24$. Solving $a^2 - 10a + 24 = 0$,we get $(a - 4)(a - 6) = 0$,so $a = 4$ or $a = 6$. The numbers are $43$ and $63$.If $b = 4$,$a(10 - a) = 4(3)(5) = 60$,which gives no integer solution for $a$.For $b \ge 5$,$b(b-1)(b+1) > 25$,and since the maximum value of $a(10-a)$ is $25$ (at $a=5$),there are no further solutions.Thus,there are $2$ such numbers: $43$ and $63$.

List-$I$	List-$II$
$(i)$ $\alpha = \beta$	$(A)$ $(ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii)$ $\alpha = 2\beta$	$(B)$ $2b^2 = 9ac$
$(iii)$ $\alpha = 3\beta$	$(C)$ $b^2 = 6ac$
$(iv)$ $\alpha = \beta^2$	$(D)$ $3b^2 = 16ac$
	$(E)$ $b^2 = 4ac$
	$(F)$ $(ac^2)^{1/3} + (a^2c)^{1/3} = b$

The number of all $2$-digit numbers $n$ such that $n$ is equal to the sum of the square of the digit in its tens place and the cube of the digit in its units place is

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