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

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

A Solution

A perfect square will be $(8k+r)^2=64k^2+16kr+r^2\equiv r^2\pmod{16}$ where $r=0,1,...,7$ .

Notice that $r^2\equiv 1,4,9,0 \pmod{16}$ .

Now $ab3c$ in base 8 is $a8^3+b8^2+3(8)+c\equiv 8+c\pmod{16}$ . It being a perfect square means $8+c\equiv 1,4,9,0 \pmod{16}$ . This leads to $c=1$ .

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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1982 AHSME Problems/Problem 26

Problem 26

A Solution

Partial and Wrong Solution