1972 AHSME Problems/Problem 11

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Problem

The value(s) of $y$ for which the following pair of equations $x^2+y^2+16=0\text{ and }x^2-3y+12=0$ may have a real common solution, are

$\textbf{(A) }4\text{ only}\qquad \textbf{(B) }-7,~4\qquad \textbf{(C) }0,~4\qquad \textbf{(D) }\text{no }y\qquad  \textbf{(E) }\text{all }y$

Solution

We can rewrite the equation $47_a = 74_b$ as $4a+7 = 7b+4$, or \[7(a+b) = 11a + 3.\] Then $7(a + b) \equiv 3 \pmod{11}$. Testing all the residues modulo 11, we find that the only solution to $7x \equiv 3 \pmod{11}$ is $x \equiv 2 \pmod{11}$, so $a + b \equiv 2 \pmod{11}$.

Now, since 7 is a digit in base $a$ and base $b$, we must have $a, b \ge 8$. We must also have $a+b \equiv 2 \pmod{11}$, so $a+b \ge 24$. We can have equality with $a=15, b=9$, so the least possible value of $a+b$ is $\boxed{24}$.