Question 1129281


 If {{{k}}} is a multiple of {{{24}}}, then {{{k}}} contains all of {{{24}}}’s primes. All you know thus far is that it contains {{{three}}}{{{ 2}}}’s and a {{{3}}}.

You also know that it is {{{not}}} a multiple of {{{16}}}. That means that it doesn’t contain the primes of {{{16}}}, which are {{{four}}} {{{2}}}'s.

That is, there are already {{{three}}} {{{2}}}’s in there, and there {{{can}}} {{{not}}} be {{{any}}} more. If there were, then {{{k}}} would be divisible by {{{16}}}.

In summary: 
{{{k}}} contains exactly {{{three}}}{{{ 2}}}’s, at {{{least}}} one {{{3}}}, and possibly some other primes you don’t know about.

If that’s the case, what numbers could divide evenly into{{{ k}}}?

we know {{{three}}}{{{ 2}}}’s make {{{8}}}
{{{8}}} divides evenly into {{{k}}}, since there are {{{three}}}{{{ 2}}}’s in {{{k}}}=>this {{{will}}}{{{ be}}} an integer 

{{{9 }}}could divide evenly into{{{ k}}}, since there {{{could}}}{{{ be}}} two {{{3}}}’s in{{{ k}}}=>this {{{might}}} be an integer 

{{{32}}} can’t divide evenly into {{{k}}}, there are {{{five}}}{{{ 2}}}’s in {{{32}}}, but there is {{{only}}}{{{ three}}}{{{ 2}}}’s in {{{k}}} 
so, {{{k/32}}} {{{DEFINITELY}}} {{{will}}} {{{not}}}{{{ be}}} an integer; consequently, this  is the right {{{answer}}}