1982 AHSME Problems/Problem 26

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 $24+c\equiv k^2 \pmod{64}$ , where $k^2$ is the perfect square.

If $k=8j$ , then $(8j)^2\equiv64j^2\equiv0 \pmod{64}$

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 $(8j+1)\equiv 64j^2+48j+9\equiv 48j+9\equiv 24+c \implies 48j\equiv 15+c$ , which clearly has no solutions for $c<10$ .

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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1982 AHSME Problems/Problem 26

Problem 26

Partial and Wrong Solution