$\lfloor \frac{n}{27} \rfloor=\lfloor \frac{n}{28} \rfloor$ [duplicate]

If $$n$$ be a non-negative integer then solve for $$n$$:$$\left\lfloor \frac{n}{27} \right\rfloor=\left\lfloor \frac{n}{28} \right\rfloor$$

My Attempt: $$\left\lfloor \frac{n}{27} \right\rfloor=\left\lfloor \frac{n}{28} \right\rfloor=k$$

I am able to get $$k<28$$ but can't proceed from here

marked as duplicate by Aqua algebra-precalculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 21 at 15:28

• Plugin many different numbers for $n$, can you observe a pattern? – Imago Apr 21 at 15:23
Hint: $${n\over 28}\geq [{n\over 28}] = [{n\over 27}] > {n\over 27}-1$$
so $$n< 27\cdot 28$$...
Another hint: Draw a graph of $$[{n\over 28}]$$ and $$[{n\over 27}]$$.