Question

The number of all 2-digit numbers n such that n is equal to the sum of the square of digit in its tens place and the cube of the digit in its unit place is
A.0
B.1
C.2
D.4

Answer 1

Hint: In this question, we have to find the number of 2-digit numbers that are possible such that the number is equal to the sum of the square of digit in its tens place and the cube of the digit in its unit place. A 2-digit number n is written as $ ab $ where $ a $ is the digit in tens place and $ b $ is the digit in the ones place so we can write $ n = 10a + b $ . We are given that the 2-digit number n is equal to the sum of the square of the digit in its tens place and the cube of the digit in its unit place, that is, $ n = {a^2} + {b^3} $ . Using this information, we can find the correct answer.

Complete step by step solution:
We have $ n = 10a + b $ and $ n = {a^2} + {b^3} $ , so we get –
\[
   \Rightarrow 10a + b = {a^2} + {b^3} \\
   \Rightarrow 10a - {a^2} = {b^3} - b \\
   \Rightarrow a(10 - a) = b({b^2} - 1) \\
   \Rightarrow a(10 - a) = b(b - 1)(b + 1) \;
 \]
At $ b = 1,\,a = 0 $ , we get a 1-digit number, so it is not considered.
At
 $
  b = 2,\,a(10 - a) = 6 \\
   \Rightarrow {a^2} - 10a + 6 = 0 \;
  $
Solving this equation, we get the value of $ a $ in decimals so it is not considered.
At
 $
  b = 3,\,a(10 - a) = 24 \\
   \Rightarrow {a^2} - 10a + 24 = 0 \\
   \Rightarrow {a^2} - 6a - 4a + 24 = 0 \\
   \Rightarrow a(a - 6) - 4(a - 6) = 0 \\
   \Rightarrow (a - 4)(a - 6) = 0 \\
   \Rightarrow a = 4,\,a = 6 \;
  $
We get 2-digit numbers, 43 and 63.
At
 $
  b = 4,\,a(10 - a) = 60 \\
   \Rightarrow {a^2} - 10a + 60 = 0 \;
  $
On solving the above equation, we get imaginary roots, so it cannot be considered.
At
 $
  b = 5,\,a(10 - a) = 120 \\
   \Rightarrow {a^2} - 10a + 120 = 0 \;
  $
We get imaginary roots on solving this equation, so it is not considered.
We don’t test further as we continue to get imaginary roots as we increase the value of b.
Thus, only 2 numbers, 43 and 63 are possible.
Hence option (C) is the correct answer.
So, the correct answer is “Option C”.

Note: In the above solution, we have obtained quadratic equations, as the highest exponent of the unknown variable quantity is 2. A polynomial equation has exactly as many solutions as the degree of the equation, so they will have 2 roots that are obtained by factorization or by using the quadratic formula.