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Difference between revisions of "1982 AHSME Problems/Problem 26"

(→‎Partial and Wrong Solution)
(→‎Partial and Wrong Solution)
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== Partial and Wrong Solution ==
 
== Partial and Wrong Solution ==
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From the definition of bases we have <math>24+c\equiv k^2 \pmod{64}</math>, where <math>k^2</math> is the perfect square.
+
From the definition of bases we have <math>k^2=512a+64b+24+c</math>, and <math>k^2\equiv 24+c \pmod{64}</math>
  
−
If <math>k=8j</math>, then <math>(8j)^2\equiv64j^2\equiv0 \pmod{64}</math>
+
If <math>k=8j</math>, then <math>(8j)^2\equiv64j^2\equiv0 \pmod{64}</math>, which makes <math>c\equiv -24\pmod{64}</math>
  
 
If <math>k=8j+1</math>, then <math>(8j+1)\equiv 64j^2+16j+1\equiv 16j+1\equiv 24+c \implies 16j\equiv 23+c</math>, which clearly can only have the solution <math>c\equiv 7 \pmod{64}</math>, for <math>j\equiv 1</math>. This makes <math>k=9</math>, which doesn't have 4 digits in base 8
 
If <math>k=8j+1</math>, then <math>(8j+1)\equiv 64j^2+16j+1\equiv 16j+1\equiv 24+c \implies 16j\equiv 23+c</math>, which clearly can only have the solution <math>c\equiv 7 \pmod{64}</math>, for <math>j\equiv 1</math>. This makes <math>k=9</math>, which doesn't have 4 digits in base 8
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If <math>k=8j+2</math>, then <math>(8j+1)\equiv 64j^2+32j+4\equiv 32j+4\equiv 24+c \implies 32j\equiv 20+c</math>, which clearly can only have the solution <math>c\equiv 12 \pmod{64}</math>, for <math>j\equiv 1</math>. <math>c</math> is greater than <math>9</math>, and thus, this solution is invalid.  
 
If <math>k=8j+2</math>, then <math>(8j+1)\equiv 64j^2+32j+4\equiv 32j+4\equiv 24+c \implies 32j\equiv 20+c</math>, which clearly can only have the solution <math>c\equiv 12 \pmod{64}</math>, for <math>j\equiv 1</math>. <math>c</math> is greater than <math>9</math>, and thus, this solution is invalid.  
  
−
If <math>k=8j+3</math>, then <math>(8j+1)\equiv 64j^2+48j+9\equiv 48j+9\equiv 24+c \implies 48j\equiv 15+c</math>, which clearly has no solutions for <math>c<10</math>.  
+
If <math>k=8j+3</math>, then <math>(8j+3)\equiv 64j^2+48j+9\equiv 48j+9\equiv 24+c \implies 48j\equiv 15+c</math>, which clearly has no solutions for <math>0\leq c<10</math>.  
  
−
Similarly, <math>k=8j+4</math>, yields no solutions
+
Similarly, <math>k=8j+4</math> yields no solutions
  
 
If <math>k=8j+5</math>, then <math>(8j+5)\equiv 64j^2+80j-1\equiv 16j-1\equiv 24+c \implies 16j\equiv 25+c</math>, which clearly can only have the solution <math>c\equiv 9 \pmod{64}</math>, for <math>j\equiv 1</math>. This makes <math>k=13</math>, which doesn't have 4 digits in base 8.  
 
If <math>k=8j+5</math>, then <math>(8j+5)\equiv 64j^2+80j-1\equiv 16j-1\equiv 24+c \implies 16j\equiv 25+c</math>, which clearly can only have the solution <math>c\equiv 9 \pmod{64}</math>, for <math>j\equiv 1</math>. This makes <math>k=13</math>, which doesn't have 4 digits in base 8.  
  
 
If <math>k=8j+6</math>, then <math>(8j+6)\equiv 64j^2+96j+1\equiv 32j+36\equiv 24+c \implies 16j\equiv 23+c</math>, which clearly can only have the solution <math>c\equiv 7 \pmod{64}</math>, for <math>j\equiv 1</math>. This makes <math>k=9</math>, which doesn't have 4 digits in base 8
 
If <math>k=8j+6</math>, then <math>(8j+6)\equiv 64j^2+96j+1\equiv 32j+36\equiv 24+c \implies 16j\equiv 23+c</math>, which clearly can only have the solution <math>c\equiv 7 \pmod{64}</math>, for <math>j\equiv 1</math>. This makes <math>k=9</math>, which doesn't have 4 digits in base 8

Revision as of 14:04, 15 April 2016

Problem 26

If the base $8$ representation of a perfect square is $ab3c$, where $a\ne 0$, then $c$ equals

$\text{(A)} 0\qquad \text{(B)}1 \qquad \text{(C)} 3\qquad \text{(D)} 4\qquad \text{(E)} \text{not uniquely determined}$

Partial and Wrong Solution

From the definition of bases we have $k^2=512a+64b+24+c$, and $k^2\equiv 24+c \pmod{64}$

If $k=8j$, then $(8j)^2\equiv64j^2\equiv0 \pmod{64}$, which makes $c\equiv -24\pmod{64}$

If $k=8j+1$, then $(8j+1)\equiv 64j^2+16j+1\equiv 16j+1\equiv 24+c \implies 16j\equiv 23+c$, which clearly can only have the solution $c\equiv 7 \pmod{64}$, for $j\equiv 1$. This makes $k=9$, which doesn't have 4 digits in base 8

If $k=8j+2$, then $(8j+1)\equiv 64j^2+32j+4\equiv 32j+4\equiv 24+c \implies 32j\equiv 20+c$, which clearly can only have the solution $c\equiv 12 \pmod{64}$, for $j\equiv 1$. $c$ is greater than $9$, and thus, this solution is invalid.

If $k=8j+3$, then $(8j+3)\equiv 64j^2+48j+9\equiv 48j+9\equiv 24+c \implies 48j\equiv 15+c$, which clearly has no solutions for $0\leq c<10$.

Similarly, $k=8j+4$ yields no solutions

If $k=8j+5$, then $(8j+5)\equiv 64j^2+80j-1\equiv 16j-1\equiv 24+c \implies 16j\equiv 25+c$, which clearly can only have the solution $c\equiv 9 \pmod{64}$, for $j\equiv 1$. This makes $k=13$, which doesn't have 4 digits in base 8.

If $k=8j+6$, then $(8j+6)\equiv 64j^2+96j+1\equiv 32j+36\equiv 24+c \implies 16j\equiv 23+c$, which clearly can only have the solution $c\equiv 7 \pmod{64}$, for $j\equiv 1$. This makes $k=9$, which doesn't have 4 digits in base 8

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