Inspired by Rajen Kapur

?.... um... $\sqrt {20160}=141.986$ , so try ${142}^{2}=20164$ ?

All right, this problem has generated numerous comments. Let's generalize a bit. Consider a number $2016$ followed by $a$ digits. Then we examine, for each $a$ , how many $n$ can there be. $n$ has to satisfy the following

$\sqrt { 2016\cdot { 10 }^{ a } } <n<\sqrt { 2017\cdot { 10 }^{ a } }$

which can be rewritten as

$\sqrt { 2016 } <n\cdot { 10 }^{ -\frac { 1 }{ 2 } a }<\sqrt { 2017 }$

or

$44.899...<n\cdot { 10 }^{ -\frac { 1 }{ 2 } a }<44.911...$

which means for a square of an even number of digits starting with digits $2016$ , the unique answer is always $449$ followed by zeroes. Likewise, the first expression can be rewritten as

$(\sqrt { 2016\cdot { 10 } } <n\cdot { 10 }^{ -\frac { 1 }{ 2 } \left( a-1 \right) }<\sqrt { 2017\cdot { 10 } } )$

or

$141.986...<n\cdot { 10 }^{ -\frac { 1 }{ 2 } \left(a-1\right) }<142.021...$

which means for a square of an odd number of digits starting with $2016$ , the unique answer is always $142$ followed by zeroes. See the following to see the uniqueness of $142$ and $449$

$20164={142}^{2}$
$201601={449}^{2}$
$2016400={1420}^{2}$
$20160100={4490}^{2}$

I chosed to use a Javascript algorithm for this:

function findSmallestPerfectSquare() { var i = 1; while (true) { var tmp = i*i; if (tmp.toString().substring(0,4) === "2016") { return tmp; } else { i++; } } } var smallestPerfSquare = findSmallestPerfectSquare(); console.log(smallestPerfSquare);

Which will generate perfect squares and then check if the perfect squares' first 4 numbers is 2016. If not proceed to next perfect square. Returns: 20164 after 10ms (not fast I know)

well, a square of anumber has in its unit place a

• 1 if the the number has 1 in its unit place. but

• 4 if the number has 2 in its unit place. Thus, 20164 is the number.