So squarish

My solution is not elegant! The number 20xyz16 is a perfect square, so the number squared must be between 2,000,000 and 2,100,000 which means that that number TO SQUARE must be between 1414.2 and 1449.1. 2,0xy,z16 is a multiple of 4 (not necessarily a multiple of 16) so the square root of this must be even. Also the square ends in 6, so the square root cannot end in 0, 2 or 8 but must end in 4 or 6. There are only 7 numbers STRICTLY between 1414.2 and 1449 which obey these conditions (i.e. not 1414 - too small). And trial and error (something that I don't like, and think that someone else will do better?!) gives that 1446^2 = 2,090,916 is the only number that obeys these few conditions. So the 3 digit answer is 909. Regards, David

Let $a^2$ be a square of an integer ending with $16$ , then it's obvious that:

$a^2 - 16 = (a-4)(a+4)$

Divisible by $100$ , so $a = 2b$ and $(b-2)(b+2)$ is divisible by 25.

Then further it must be true, that $b = 25n \pm 2$ and $a = 50n \pm 4$ , because:

$1404^2 < (1.414 \times 1000)^2 < (1000\sqrt2)^2 = 2000000$ .

$1454^2 > 1450^2 = 2102500 > 2100000$ .

Hence $a = 1446$ which gives us $a^2 = 20\boxed{909}16$

Because I know that $14^2=196$ and $15^2=225$ ,

so $1400^2=1960000$ and $1500^2=2250000$ .

I tried $1450^2$ and get $2102500$ .

So I am sure that the 7-digit integer is $1444^2$ or $1446^2$ . (This is because only a number with last digit 4 or 6 brings last digit to 6 after squaring it)

I test the last two digit and found $44^2=1936$ and $46^2=2116$ .

Therefore the 7-digit integer is $1446^2=2090916$ , the missing digit is $\boxed{909 }$ .

$20[][][]16 = n^2$ well 16 is a pretty number so I'm gonna write this as a sum of numbers.

$20[][][]00 + 16 = a^2 + 2ab + b^2$ where $n = a+b$ .

If $b^2 = 16$ then $20[][][]00 = a^2 + 2*4a$ .

If $20[][][]00 = 20[][][]*100$ then $\frac{a^2}{100} + \frac{2a}{25}$ has to be natural numbers since $20[][][]$ is probably a natural number too.

Then I should be able to factorize $a$ to $25r$ . Now we have $\frac{25r^2}{4} + 2r$ .

Following the same logic then I should be able to factorize $r$ to $2t$ so that $\frac{25r^2}{4} + 2r$ becomes $25t^2 + 4t$ .

Well, I have no idea what I can do now so I'm gonna estimate it. Let us solve for $t$ where $25t^2 + 4t = \frac{20000+20999}{2}$ and after that just round off whatever we get.

If we still don't find the correct solution then we are going to add one or remove one from $t$ until $25t^2 + 4t < 20000$ or $25t^2 + 4t > 20999$ .

$t = -29$ or $t = 29$ . Let us put the solutions back into the function $25(-29)^2 + 4*(-29) = 20\boxed{909}$ .

$\sqrt{20\boxed{909}16} = 1446$ and would you look at that. No wonder why 85% of everyone managed to solve this problem, jk this was kinda hard.

. (WXYZ)^2=20•••16

Which means that the first digit from left has the form of: W^2+ a carry = 2 This has two possibilities, either w^2=1 & the carry=1 Or W^2=2 & the carry=0; which is rejected since W is an integer ... So W=1. Now for Z & Y; if we take a look at the first digit from right, which is 6, then there are two possibilities, either Z= 4 or Z=6, for the first case of Z=4, we can find a value of Y such that: (4) (2) (Y)+ ((4^2-(4^2%10))/10= a number with 1 as the first digit from the right, then 8 Y= a number with zero as a first digit from the right... So Y=5(initially we can test the number 1X54 for different values of X and conclude this is not a solution). For the second case of Z=6, Y will have a form: (6) (2)*(Y)+(6^2-(6^2%10))/10=a number with 1 as the first digit from the right ... 12Y=a number with 8 as the first digit from right, solving this will lead to two possibilities, either Y=4(testing the number 1X46 for different values of X will show that this is a solution, and I will prove this using equations later) or Y=9(initially we can test the number 1X96 by the calculator for different values of X ,,, so this is not a solution)

To find a Value of X ... We use the following equations: Noting that the second digit from the left has a value of zero and result with a carry of 1 means that the summation which result in the digit of zero=10, so... 2X+carry3=10 ...(1)

Carry3=((X^2+8+carry2)-(X^2+8+carry2)%10)/10 ...(2)

Carry2=((8X+13+carry1)-(8X+13+carry1)%10)/10 ...(3)

Carry1=((12X+11)-(12X+11)%10)/10 ... (4)

Solving for different values of X ;where X between 0 & 9, & X satisfies equation #1, yields to X=4

So the result is (WXYZ)^2=(1446)^2=2090916