가우스 기호가 포함된 합을 11로 나눈 나머지

$n$을 $11$로 나누었을 때 몫과 나머지를 각각 $Q , R$이라 하면 $n = 11 Q + R$ (단, $0 \leq R \leq 10$인 정수) 이므로 $\displaystyle \sum\limits _{n = 1} ^{2004} \left[ \frac{n ^{2}}{11} \right]$$\displaystyle = \sum\limits _{n = 0} ^{2004} \left[ \frac{n ^{2}}{11} \right]$$\displaystyle = \sum\limits _{n = 0} ^{2001} \left[ \frac{n ^{2}}{11} \right] + \sum\limits _{n = 2002} ^{2004} \left[ \frac{n ^{2}}{11} \right]$ $\displaystyle = \sum\limits _{Q = 0} ^{181} \sum\limits _{R = 0} ^{10} \left[ \frac{( 11 Q + R ) ^{2}}{11} \right] + \sum\limits _{n = 2002} ^{2004} \left[ \frac{n ^{2}}{11} \right]$ $\displaystyle = \sum\limits _{Q = 0} ^{181} \sum\limits _{R = 0} ^{10} \left[ \frac{11 ^{2} Q ^{2} + 2 \cdot 11 \cdot Q \cdot R + R ^{2}}{11} \right] + \sum\limits _{n = 2002} ^{2004} \left[ \frac{n ^{2}}{11} \right]$ $\displaystyle = \sum\limits _{Q = 0} ^{181} \sum\limits _{R = 0} ^{10} \left\{ 11 Q ^{2} + 2 \cdot Q \cdot R + \left[ \frac{R ^{2}}{11} \right] \right\} + \sum\limits _{n = 2002} ^{2004} \left[ \frac{n ^{2}}{11} \right]$ $\displaystyle = \sum\limits _{Q = 0} ^{181} \left\{ 11 ^{2} Q ^{2} + 10 \cdot 11 Q + \sum\limits _{R = 0} ^{10} \left[ \frac{R ^{2}}{11} \right] \right\} + \sum\limits _{n = 2002} ^{2004} \left[ \frac{n ^{2}}{11} \right] \cdots$(1) 여기서 $\displaystyle \sum\limits _{R = 0} ^{10} \left[ \frac{R ^{2}}{11} \right] = \left[ \frac{0 ^{2}}{11} \right] + \left[ \frac{1 ^{2}}{11} \right] + \cdots + \left[ \frac{10 ^{2}}{11} \right] = 31$ 이므로 식 (1)은 $\displaystyle = \sum\limits _{Q = 0} ^{181} \left\{ 11 ^{2} Q ^{2} + 10 \cdot 11 Q + 31 \right\} + \sum\limits _{n = 2002} ^{2004} \left[ \frac{n ^{2}}{11} \right] \cdots$(2) 따라서, 구하고자 하는 값은 $\displaystyle 181 \cdot 31 + \sum\limits _{n = 2002} ^{2004} \left[ \frac{n ^{2}}{11} \right] = 31 \cdot ( 6 \cdot 11 + 5 ) + \sum\limits _{n = 2002} ^{2004} \left[ \frac{n ^{2}}{11} \right]$ 을 $11$로 나누었을 때의 나머지와 같다. $\displaystyle \sum\limits _{n = 2002} ^{2004} \left[ \frac{n ^{2}}{11} \right] = \sum\limits _{R = 0} ^{2} \left[ \frac{( 11 \cdot 182 + R ) ^{2}}{11} \right]$ 의 관계를 이용하여 구하면 나머지는 $2$이다. 【다른 풀이 1】 $\displaystyle \left[ \frac{n ^{2}}{11} \right]$은 $1$에서 $n ^{2}$까지의 11의 배수의 개수이다. $\displaystyle \sum\limits _{{n} = 1} ^{2004} \left[ \frac{n ^{2}}{11} \right] = \sum\limits _{{n} = 0} ^{2004} \left[ \frac{n ^{2}}{11} \right]$이다. ⅰ) $n = 0 , 1 , 2 , \cdots , 10 ( = 11 \times 0 + 10 )$ 을 고려하면, $\displaystyle \sum\limits _{{n} = 0} ^{10} \left[ \frac{n ^{2}}{11} \right] = 0 + 0 + 0 + 0 + 1 + 2 + 3 + 4 + 5 + 7 + 9 = 31$이다. 이를 다시 $11$로 나누어 보면, $31 = 11 \times 2 + 9$이다. ⅱ) $n = 11 , 12 , 13 , \cdots , 21 ( = 11 \times 1 + 10 )$ 을 고려하면, $n = 11 \times 1 + r$ (단,$0 \leq r \leq 10$ ) 이라고 놓으면, $\displaystyle \sum\limits _{{n} = 11} ^{21\left[ \frac{n ^{2}}{11} \right]} = \sum\limits _{{r} = 0} ^{10\left[ \frac{{( 11 \times 1 + r} ) ^{2}}{11} \right]}$ $\displaystyle = \sum\limits _{{r} = 0} ^{10} \left[ \frac{{11 ^{2} \times 1 ^{2} + 2 \times 11 \times 1 \times r + r ^{2}}}{11} \right]$ $\displaystyle = \sum\limits _{{r} = 0} ^{10} \left[ 11 \cdot 1 ^{2} + 2 \times 1 \times r + \frac{{r ^{2}}}{11} \right]$ $\displaystyle = \sum\limits _{{r} = 0} ^{10} \left\{ 11 \cdot 1 ^{2} + 2 \times 1 \times r + \left[ \frac{{r ^{2}}}{11} \right] \right\}$ $\displaystyle = 11 \times 11 + 11 \times 2 r + \sum\limits _{{r} = 0} ^{10} \left[ \frac{{r ^{2}}}{11} \right]$ $= 11 ( 11 + 2 r ) + 31 = 11 ( 11 + 2 r + 22 ) + 9$ 따라서, $\displaystyle \sum\limits _{{n} = 11} ^{21\left[ \frac{n ^{2}}{11} \right]}$을 $11$로 나눈 나머지는 $9$이다. 이런 방법으로 ⅲ) $n = 22 , 23 , 24 , \cdots , 32 ( = 11 \times 2 + 10 )$ 을 고려하면, $\displaystyle \sum\limits _{{n} = 22} ^{32\left[ \frac{n ^{2}}{11} \right]}$을 $11$로 나눈 나머지도 $9$이다. 이러한 방법으로 계속 계산하면 ⅳ) $n = 1991 , 1992 , 1993 , \cdots , 2001 ( = 11 \times 181 + 10 )$ 을 고려하면,$\displaystyle \sum\limits _{{n} = 1991} ^{2001\left[ \frac{n ^{2}}{11} \right]}$을 $11$로 나눈 나머지도 $9$이다. 따라서, $\displaystyle \sum\limits _{{n} = 1} ^{2001} \left[ \frac{n ^{2}}{11} \right] = \sum\limits _{{n} = 0} ^{2001} \left[ \frac{n ^{2}}{11} \right]$ 를 $11$로 나눈 나머지는 $182 \times 9$를 $11$로 나눈 나머지이므로 $182 \times 9 = ( 11 \times 16 + 6 ) \times 9 = 11 \times 16 \times 9 + 54 = 11 ( 16 \times 9 + 4 ) + 10$ 에서 나머지는 $10$이다.$\cdots ( 1 )$ 또, $\displaystyle \sum\limits _{{n} = 2002} ^{2004} \left[ \frac{n ^{2}}{11} \right]$를 $11$로 나눈 나머지를 구해보면, $\displaystyle \sum\limits _{{n} = 2002} ^{2004} \left[ \frac{n ^{2}}{11} \right] = \left[ \frac{2002 ^{2}}{11} \right] + \left[ \frac{2003 ^{2}}{11} \right] + \left[ \frac{2004 ^{2}}{11} \right]$ $\displaystyle = \left[ \frac{( 11 \times 182 ) ^{2}}{11} \right] + \left[ \frac{( 11 \times 182 + 1 ) ^{2}}{11} \right] + \left[ \frac{( 11 \times 182 + 2 ) ^{2}}{11} \right]$ $\displaystyle = 11 \times 182 ^{2} + \left[ \frac{( 11 \times 182 ) ^{2} + 2 \times 11 \times 182 \times 1 + 1}{11} \right]$ $\displaystyle + \left[ \frac{( 11 \times 182 ) ^{2} + 2 \times 11 \times 182 \times 2 + 4}{11} \right]$ $\displaystyle = 11 \times 182 ^{2} + \left[ {11 \times 182 ^{2} + 2 \times 182 \times 1 + \frac{1}{11}} \right]$ $\displaystyle + \left[ {11 \times 182 ^{2} + 2 \times 182 \times 2 + \frac{4}{11}} \right]$ $\displaystyle = 11 \times 182 ^{2} + {11 \times 182 ^{2} + 2 \times 182 \times 1 + \left[ \frac{1}{11} \right. ]}$ $\displaystyle + {11 \times 182 ^{2} + 2 \times 182 \times 2 + \left[ \frac{4}{11} \right. ]}$ $= 3 \times 11 \times 182 ^{2} + {2 \times 182 + 0} + {4 \times 182 + 0}$ $= 3 \times 11 \times 182 ^{2} + 6 \times 182$ $= 3 \times 11 \times 182 ^{2} + 6 \times ( 11 \times 16 + 6 )$ $= 11 ( 3 \times 182 ^{2} + 6 \times 16 ) + 36$ $= 11 ( 3 \times 182 ^{2} + 6 \times 16 + 3 ) + 3$ 따라서,$11$로 나눈 나머지는 $3$이다.$\cdots ( 2 )$ $\therefore ( 1 ) , ( 2 )$에서 $\displaystyle \sum\limits _{{n} = 1} ^{2004} \left[ \frac{n ^{2}}{11} \right] = \sum\limits _{{n} = 0} ^{2004} \left[ \frac{n ^{2}}{11} \right]$를 $11$로 나눈 나머지는 $10 + 3 = 13$ 을 $11$로 나눈 나머지이므로 정답은$2$이다. 【다른 풀이 2】 : mod 11 이용 $\displaystyle \sum\limits _{{n} = 1} ^{2004} \left[ \frac{n ^{2}}{11} \right] = \sum\limits _{{n} = 0} ^{2004} \left[ \frac{n ^{2}}{11} \right]$ $\displaystyle \equiv \sum\limits _{{k} = 0} ^{181} \left\{ \sum\limits _{{l} = 0} ^{10} \left[ \frac{( 11 k + l ) ^{2}}{11} \right] \right\} + \sum\limits _{{n} = 2002} ^{2004} \left[ \frac{n ^{2}}{11} \right]$ ( $mod 11$ ) $\displaystyle \equiv \sum\limits _{{k} = 0} ^{181} \left\{ \sum\limits _{{l} = 0} ^{10} \left[ \frac{( 11 ^{2} k ^{2} + 2 \times 11 k l + l ^{2} )}{11} \right] \right\} + \left[ \frac{( 11 \times 182 + 0 ) ^{2}}{11} \right]$ $\displaystyle + \left[ \frac{( 11 \times 182 + 1 ) ^{2}}{11} \right] + \left[ \frac{( 11 \times 182 + 2 ) ^{2}}{11} \right]$ $\displaystyle \equiv \sum\limits _{{k} = 0} ^{181} \left\{ \sum\limits _{{l} = 0} ^{10} \left( 11 k ^{2} + 2 kl + \left[ \frac{l ^{2}}{11} \right] \right) \right\} + 0 + 1 + 2$ $\displaystyle \equiv \sum\limits _{{k} = 0} ^{181} \left\{ 11 ^{2} k ^{2} + 55 k + \sum\limits _{{l} = 0} ^{10} \left[ \frac{l ^{2}}{11} \right] \right\} + 3$ $\displaystyle \equiv 0 + 0 + \sum\limits _{{k} = 0} ^{181} \left\{ \sum\limits _{{l} = 0} ^{10} \left[ \frac{l ^{2}}{11} \right] \right\} + 3$ $\displaystyle \equiv \sum\limits _{{k} = 0} ^{181} ( 0 + 0 + 0 + 0 + 1 + 2 + 3 + 4 + 5 + 7 + 9 ) + 3$ $\equiv 182 \times 31 + 3 \equiv 6 \times 9 + 3 \equiv 54 + 3 \equiv 10 + 3 \equiv 2$