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    <p>You can use bitwise and for your operations</p> <pre><code>declare @i int declare @x int declare @cnt int select @i=2147483647 -- 1111111 11111111 11111111 11111111 declare @t table (a int) insert into @t values( 3),(10),(524290); Select @i= (@i &amp; a) from @t -- just for fun an output set @x=1 set @cnt=0 While @cnt&lt;31 begin Print Case when @x &amp; @i &lt;&gt; 0 then 'X' else ' ' end +' ' + Cast(@x as Varchar(12)) Set @cnt=@cnt + 1 if @cnt&lt;31 Set @x=@x*2 end </code></pre> <p>or with a nicer output</p> <pre><code>declare @i int declare @x int declare @cnt int select @i=2147483647 -- 1111111 11111111 11111111 11111111 Declare @ref table(ref int) set @x=1 set @cnt=0 While @cnt&lt;31 begin insert into @ref Values(@x) Set @cnt=@cnt + 1 if @cnt&lt;31 Set @x=@x*2 end declare @t table (a int) insert into @t values( 3),(10),(524290); Select @i= (@i &amp; a) from @t Select * from @ref where ref&amp;@i&lt;&gt;0 </code></pre> <p>as answer to your comment</p> <pre><code>declare @i int declare @x int declare @cnt int select @i=2147483647 -- 1111111 11111111 11111111 11111111 Declare @ref table(ref int) set @x=1 set @cnt=0 While @cnt&lt;31 begin insert into @ref Values(@x) Set @cnt=@cnt + 1 if @cnt&lt;31 Set @x=@x*2 end declare @t table (a int) insert into @t values( 3),(5),(9),(17),(33),(65),(128); Select @i= (@i &amp; a) from @t Select a,Count(*) as BitCount from @ref r join @t t on t.a &amp; r.ref&lt;&gt;0 group by a Select ref,Count(*) as OCC from @ref r join @t t on t.a &amp; r.ref&lt;&gt;0 group by ref Select ref,(Select count(*) from @t where a &amp; r.ref&lt;&gt;0) as OCC from @ref r </code></pre>
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    1. POcalculating bit flag intersection for congruency
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    1. COThat’s a nice approach with `SELECT @i = (@i & a) FROM @t` doing the magic. At first glance, it wasn’t obvious to me what’s actually assigned to `@i`, though it truly yields a decimal representation of bits that are set in each of the integers in the `@t` table. Now imagine a set like `3,5,9,17,33,65,128`. There’s `2` set in 7 out of 8 values, but the solution above won’t return anything else but `0`.
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