Question

A leap of coconuts is divided into groups of $2,3$ and $5$ and each time one coconut is left over. the least number of coconuts in the leap is

• A. $11$
• B. $21$
• C. $31$
• D. $41$

Answer 1

The correct option is C $31$

The number which is divisible by $2,3,5$ will be LCM of these which is $30$ now on dividing we are left with remainder 1 in each case
$\therefore numberis30+1=31$