Question

The sum of the odd days of two consecutive years is

Answer 1

$1</p><p> ordinary year </p><p>=</p><p>365</p><p> days </p><p>=</p><p>1</p><p> odd day</p><p>1</p><p> leap year </p><p>=</p><p>366</p><p> days </p><p>=</p><p>2</p><p> odd days</p><p>100</p><p> years </p><p>=</p><p> (</p><p>76</p><p> ordinary years + </p><p>24</p><p> leap years) </p><p>=</p><p>(</p><p>76</p><p>×</p><p>1</p><p>+</p><p>24</p><p>×</p><p>2</p><p>)</p><p> odd days</p><p>=</p><p>124</p><p> odd days </p><p>=</p><p>5</p><p> odd days.</p><p>200</p><p> years </p><p>=</p><p>(</p><p>5</p><p>×</p><p>2</p><p>)</p><p>=</p><p>10</p><p>=</p><p>3</p><p> odd days.</p><p>300</p><p> years </p><p>=</p><p>(</p><p>5</p><p>×</p><p>3</p><p>)</p><p>=</p><p>15</p><p>=</p><p>1</p><p> odd day.</p><p>400</p><p> years</p><p>=</p><p>(</p><p>5</p><p>×</p><p>4</p><p>+</p><p>1</p><p>)</p><p>=</p><p>21</p><p>=</p><p>0</p><p>odd days.</p><p>Similarly, odd days in every </p><p>4</p><p>th</p><p> century (</p><p>400</p><p>,</p><p>800</p><p>,</p><p>1200</p><p> etc.) </p><p>=</p><p>0</p><p>Odd \: days \: corresponding \: to \: every \: month \: is \: provided \: below.$