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

Options

A. 0
B. 1
C. 2
D. 4

Explanation

To solve the problem, we need to find all two-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.

Let's denote:
- \( a \) as the digit in the tens place,
- \( b \) as the digit in the units place.

Thus, we can express the two-digit number \( n \) as:
\[ n = 10a + b \]

According to the problem, we have:
\[ n = a^2 + b^3 \]

Substituting the expression for \( n \) into the equation gives:
\[ 10a + b = a^2 + b^3 \]

Rearranging this equation, we get:
\[ a^2 - 10a + b^3 - b = 0 \]

This is a quadratic equation in terms of \( a \):
\[ a^2 - 10a + (b^3 - b) = 0 \]

To find valid values for \( a \), we need the discriminant of this quadratic equation to be non-negative:
\[ D = b^3 - b + 100 \geq 0 \]

Next, we will check for each digit \( b \) from 0 to 9 (since \( b \) is a digit in the units place) and find the corresponding values of \( a \).

### Step 1: Check values of \( b \)

1. **For \( b = 0 \)**:
\[ D = 0^3 - 0 + 100 = 100 \quad (\text{valid}) \]
Solving for \( a \):
\[ a^2 - 10a + 0 = 0 \implies a(a - 10) = 0 \implies a = 0 \text{ or } 10 \quad (\text{not valid, as } a \text{ must be a digit}) \]

2. **For \( b = 1 \)**:
\[ D = 1^3 - 1 + 100 = 100 \quad (\text{valid}) \]
Solving for \( a \):
\[ a^2 - 10a + 0 = 0 \implies a(a - 10) = 0 \implies a = 0 \text{ or } 10 \quad (\text{not valid}) \]

3. **For \( b = 2 \)**:
\[ D = 2^3 - 2 + 100 = 106 \quad (\text{valid}) \]
Solving for \( a \):
\[ a^2 - 10a + 6 = 0 \]
The discriminant is:
\[ D = (-10)^2 - 4 \cdot 1 \cdot 6 = 100 - 24 = 76 \quad (\text{valid}) \]
The solutions for \( a \):
\[ a = \frac{10 \pm \sqrt{76}}{2} \quad (\text{not integer solutions}) \]

4. **For \( b = 3 \)**:
\[ D = 3^3 - 3 + 100 = 104 \quad (\text{valid}) \]
Solving for \( a \):
\[ a^2 - 10a + 24 = 0 \]
The discriminant is:
\[ D = 100 - 96 = 4 \quad (\text{valid}) \]
The solutions for \( a \):
\[ a = \frac{10 \pm 2}{2} \implies a = 6 \text{ or } 4 \]
Thus, valid numbers are \( 63 \) and \( 43 \).

5. **For \( b = 4 \)**:
\[ D = 4^3 - 4 + 100 = 108 \quad (\text{valid}) \]
Solving for \( a \):
\[ a^2 - 10a + 60 = 0 \]
The discriminant is:
\[ D = 100 - 240 = -140 \quad (\text{not valid}) \]

6. **For \( b = 5 \)**:
\[ D = 5^3 - 5 + 100 = 120 \quad (\text{valid}) \]
Solving for \( a \):
\[ a^2 - 10a + 120 = 0 \]
The discriminant is:
\[ D = 100 - 480 = -380 \quad (\text{not valid}) \]

7. **For \( b = 6 \)**:
\[ D = 6^3 - 6 + 100 = 210 \quad (\text{valid}) \]
Solving for \( a \):
\[ a^2 - 10a + 210 = 0 \]
The discriminant is:
\[ D = 100 - 840 = -740 \quad (\text{not valid}) \]

8. **For \( b = 7 \)**:
\[ D = 7^3 - 7 + 100 = 342 \quad (\text{valid}) \]
Solving for \( a \):
\[ a^2 - 10a + 342 = 0 \]
The discriminant is:
\[ D = 100 - 1368 = -1268 \quad (\text{not valid}) \]

9. **For \( b = 8 \)**:
\[ D = 8^3 - 8 + 100 = 508 \quad (\text{valid}) \]
Solving for \( a \):
\[ a^2 - 10a + 508 = 0 \]
The discriminant is:
\[ D = 100 - 2032 = -1932 \quad (\text{not valid}) \]

10. **For \( b = 9 \)**:
\[ D = 9^3 - 9 + 100 = 730 \quad (\text{valid}) \]
Solving for \( a \):
\[ a^2 - 10a + 730 = 0 \]
The discriminant is:
\[ D = 100 - 2920 = -2820 \quad (\text{not valid}) \]

### Conclusion
The valid two-digit numbers \( n \) that satisfy the condition are \( 43 \) and \( 63 \). Therefore, the total count of such numbers is:

\[ \text{Total count} = 2 \]