Difference between revisions of "1972 AHSME Problems/Problem 11"

Latest revision as of 13:37, 4 February 2022

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

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

Because x² + y² + 16 = 0 has no real solutions, ∀ sets containing x² + y² + 16 = 0, no real solutions may exist.

∴ the solution is $\fbox{D}$

– TylerO_1.618

@@ Line 3: / Line 3: @@
 <math>\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</math>
+==Solution==
+Because x<sup>2</sup> + y<sup>2</sup> + 16 = 0 has no real solutions, &#8704; sets containing x<sup>2</sup> + y<sup>2</sup> + 16 = 0, no real solutions may exist.
+&#8756; the solution is <math>\fbox{D}</math>
-==Solution==
-We can rewrite the equation <math>47_a = 74_b</math> as <math>4a+7 = 7b+4</math>, or <cmath> 7(a+b) = 11a + 3. </cmath>
-Then <math>7(a + b) \equiv 3 \pmod{11}</math>. Testing all the residues modulo 11, we find that the only solution to <math>7x \equiv 3 \pmod{11}</math> is <math>x \equiv 2 \pmod{11}</math>, so <math>a + b \equiv 2 \pmod{11}</math>.
-Now, since 7 is a digit in base <math>a</math> and base <math>b</math>, we must have <math>a, b \ge 8</math>. We must also have <math>a+b \equiv 2 \pmod{11}</math>, so <math>a+b \ge 24</math>. We can have equality with <math>a=15, b=9</math>, so the least possible value of <math>a+b</math> is <math>\boxed{24}</math>.
+– TylerO_1.618