Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(just grabbed solution off aops website - not my work, but work of some aops person)
 
Line 1: Line 1:
 +
==Problem==
 +
The value(s) of <math>y</math> for which the following pair of equations <math>x^2+y^2+16=0\text{ and }x^2-3y+12=0</math> may have a real common solution, are
 +
 +
<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==
 
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>
 
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>.
 
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>.
 
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>.

Revision as of 20:54, 22 June 2021

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 $\boxed{24}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1972_AHSME_Problems/Problem_11&oldid=156641"