Floor function and 2016

Algebra Level 5

$\large \left\lfloor \frac { { 1 }^{ 2 } }{ 2016 } \right\rfloor ,\left\lfloor \frac { { 2 }^{ 2 } }{ 2016 } \right\rfloor ,\left\lfloor \frac { { 3 }^{ 2 } }{ 2016 } \right\rfloor ,\left\lfloor \frac { { 4 }^{ 2 } }{ 2016 } \right\rfloor , \ldots ,\left\lfloor \frac { { 2016 }^{ 2 } }{ 2016 } \right\rfloor$

How many distinct integers are there in the sequence above?

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

The answer is 1513.

1 solution

Choi Chakfung
May 6, 2016

$\frac { { n }^{ 2 } }{ 2016 } -\frac { { (n-1) }^{ 2 } }{ 2016 } =\frac { 2n+1 }{ 2016 } \\ \frac { 2n+1 }{ 2016 } <1\quad when\quad n<1007.5\quad \\ \frac { { 1007 }^{ 2 } }{ 2016 } =503.000496..>503\\ Therefore\quad for\quad first\quad 1007\quad th\quad terms\quad ,\\ \\ the\quad number\quad include\quad 0-503\quad which\quad contain\quad 504\quad different\quad number.\\ for\quad 1008-2016\quad th\quad terms,\quad \frac { 2n+1 }{ 2016 } >1\quad when\quad n>1007.5\quad \\ every\quad number\quad will\quad larger\quad than\quad the\quad previous\quad number\quad at\quad least\quad one\quad \\ such\quad as\quad \frac { { 1008 }^{ 2 } }{ 2016 } =504>\frac { { 1007 }^{ 2 } }{ 2016 } \\ for\quad from\quad 1008-2016\quad ,\quad there\quad are\quad 2016-1008+1=\quad 1009\quad terms\\ \\ Therefore,1009+504=1513$

Floor function and 2016

The answer is 1513.

1 solution

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