Find the least number of Easter that could be in the basket

Question

Eggs in a basket are removed 2 at a time. Then 1 will remain in Basket.  Eggs 3 at a time , then 2 remain.  If eggs are removed four, five , or six at time, then 3, 4, and 5 gives eggs will remain in the basket respectively.  But if the eggs are removed seven at time, then no eggs will be left over. 

Answer 1

 

Let "E" = number of eggs in basket.

 

NOTE:  taking "x" at a time can be thought of as "dividing by x"

 

Note the pattern for the remainder ...

   

    When dividing by 2, remainder is 1.  => E is odd

    When dividing by 3, remainder is 2.

    When dividing by 4, remainder is 3,

    When dividing by 5, remainder is 4   => E ends in 4 or 9 ... exclude ending in 4 because E is odd

    When dividing by 6, remainder is 5

    When dividing by 7, remainder is 0   => E is a whole number multiple of 7

 

From the above we can conclude the following

 

    Number is a whole number multiple of 7.

 

Multiples of 7 (odd multiples) that end in 9 include:

 

    49, 119

 

When testing these out, 119 is the only one that passes.

 

There are 119 eggs in the basket.