Question 560031
Start with any number. Call it {{{ n }}}
Then write down the number {{{ n }}}
and the following 3 numbers this way:
{{{ n }}}
{{{ n + 1 }}}
{{{ n + 2 }}}
{{{ n + 3 }}}
Now add these up. You get
{{{ 4n + 6 }}}
{{{ 4n }}} is divisible by {{{4}}}, then when 
{{{6}}} is divided by {{{4}}}, you always 
get a remainder of {{{2}}}.
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Here's an example:
Suppose {{{ n = 17 }}}
{{{ 17 }}}
{{{ 17 + 1 }}}
{{{ 17 + 2 }}}
{{{ 17 + 3 }}}
Adding these 4 consecutive numbers,
{{{ 4*17 + 6 }}}
If I divide 4 {{{ 17}}} 's by {{{ 4 }}}, I get {{{ 17 }}}
Now I divide the {{{ 6 }}} by {{{ 4 }}}, and I get {{{1}}}
with {{{ 2 }}} left over, so my answer is
{{{ 17 + 1 + 2/4 }}} and that equals
18 1/2, which is not a whole number of times
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Since {{{17}}} was any number, this works for all numbers