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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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 $63$.

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

Similar Questions

If the roots of the equation $3x^2 + 4kx + 3 = 0$ are non-real,then $k$ lies in the interval

The value of $k$ for which the equation $2x^2 - kx + x + 8 = 0$ has equal and real roots is:

The roots of the equation $|x - 2|^2 + |x - 2| - 6 = 0$ are:

If the roots of $4x^2 + px + 9 = 0$ are equal,then the absolute value of $p$ is:

The point $(t^2 + 2t + 5, 2t^2 + t - 2)$ lies on the line $x + y = 2$ for

Vedclass Test Series

Exam Paper Generator

Online Exam Module