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| | <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> | | <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> |
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| − | ==Solution==
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| − | 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>
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| − | 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>.
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| − | 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>.
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Revision as of 20:57, 22 June 2021
Problem
The value(s) of 
for which the following pair of equations 
may have a real common solution, are
for which the following pair of equations 
may have a real common solution, are
