Let's denote the two digits of the figure as \( a \) and \( b \). Then, the original figure can be represented as the two-digit number \( 10a + b \).
According to the problem, if we add both digits (\( a + b \)), square that sum, and then add 9, we should have the same result as interchanging the digits (which gives the number \( 10b + a \)).
We can express that mathematically as follows:
\[ (a + b)^2 + 9 = 10b + a \]
Now, let's simplify this equation step by step.
- Expand the left side:
\[ a^2 + 2ab + b^2 + 9 = 10b + a \]
- Rearranging gives us:
\[ a^2 + 2ab + b^2 - 10b - a + 9 = 0 \]
This is a quadratic in terms of \( a \):
\[ a^2 + (2b - 1)a + (b^2 - 10b + 9) = 0 \]
To find integer values for \( a \) and \( b \), we can apply the quadratic formula \( a = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \):
Here, \( A = 1 \), \( B = 2b - 1 \), and \( C = b^2 - 10b + 9 \).
The discriminant (inside the square root) should be a perfect square:
\[ \Delta = (2b - 1)^2 - 4(1)(b^2 - 10b + 9) \]
Calculating \(\Delta\):
\[ \Delta = (4b^2 - 4b + 1) - (4b^2 - 40b + 36) \] \[ = 4b - 35 \]
For \(\Delta\) to be a perfect square, \( 4b - 35 \) must be a non-negative perfect square.
Let \( k^2 = 4b - 35 \).
Rearranging gives:
\[ 4b = k^2 + 35 \implies b = \frac{k^2 + 35}{4} \]
Next, since \( b \) must be a digit (0 to 9), we can try different integer values of \( k\).
- \( k = 1 \): \( b = \frac{1 + 35}{4} = 9 \) → valid
- \( k = 3 \): \( b = \frac{9 + 35}{4} = 11 \) → invalid
- \( k = 5 \): \( b = \frac{25 + 35}{4} = 15 \) → invalid
- \( k = 7 \): \( b = \frac{49 + 35}{4} = 21 \) → invalid
- \( k = 0 \): \( k^2 = 0 \), \( b = \frac{35}{4} = 8.75 \) → invalid
- \( k = 2, k = 4, k = 6 \): similar results; non-valid.
From \( k = 1 \), we found \( b = 9 \).
Substituting \( b = 9 \) back to see values of \( a \):
We need to verify with \( b = 9 \): \[ (2(9) - 1)a + (9^2 - 10(9) + 9) = 0 \]
Calculating leads to: \[ 17a + 0 = 0 \rightarrow a = 0, \] which forms digit 09.
Thus, our original number is \( 09 \) or \( 9 \), and the final reflex figure is:
Final Figure Reflex: 9 → 9.
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