Question 651783

If there were {{{no}}} eggs {{{left}}}{{{ over}}} when put into groups of {{{seven}}} there {{{must}}} have been a {{{multiple}}} of {{{7}}} eggs. 


We need to find {{{n}}} where
{{{n = 7a}}}
{{{n = 6b + 1}}}
{{{n = 5c + 1}}}
{{{n = 4d + 1}}}

where {{{a}}}, {{{b}}}, {{{c}}}, and {{{ d}}} are the number of {{{full}}} {{{groups}}} when the eggs are put into groups of {{{7}}}, {{{6}}}, {{{5}}}, and {{{4}}}.

{{{If}}} there is {{{one}}}{{{ egg}}}{{{ left}}}{{{ over}}} when the eggs are put into groups of {{{6}}} there will automatically be {{{one}}}{{{ egg}}}{{{ left}}}{{{ over}}} when they are put into groups of {{{2}}} or groups of {{{3}}}. So we can forget groups of {{{2}}} and {{{3}}}.

The {{{smallest}}} number which is {{{divisible}}} by {{{5}}} and {{{6}}} is {{{30}}}, but {{{30}}} isn't divisible by {{{4}}}, so the smallest number which will divide by {{{4}}}, {{{5}}}, and {{{6}}} is {{{60}}}.

So we need

{{{n = 60e + 1}}}
but also

{{{n = 7a}}}

The easiest way to solve this is to try values {{{1}}},{{{ 2}}}, {{{3}}},{{{ 4}}}, {{{5}}},... for {{{e}}} to get values of {{{61}}}, {{{121}}}, {{{181}}}, {{{241}}}, {{{301}},{{{420}}} ... for {{{n}}}, and to pick the first of these which divides by {{{7}}}. This is {{{301}}}.

So the {{{smallest}}} number of eggs is {{{301}}}.

Adding {{{60 * 7}}} to a valid solution will also give a valid solution.

So the next solution is {{{301 + 420 = 721}}}.

All the solutions are {{{301}}}, {{{721}}}, {{{1141}}}, {{{1561}}}, {{{1981}}}, ... (i.e. add {{{420}}} each time)