Year's Beginning

For $X^2$ to have starting digits of 2017. This simply means that $2017 \times 10^n \leq X^2 < 2018 \times 10^n$

Now before taking square roots on both sides, we have to check that whether $n$ is odd or even.

Considering both the cases:

$n \text{ odd: } 142.021125189 \ldots \times 10^k \leq X < 142.056326857 \ldots \times 10^k$

$n \text{ even: } 44.9110231458 \ldots \times 10^k \leq X < 44.9221548904\ldots \times 10^k$

When $n$ is odd, as we vary $k$ , the smallest integer $X$ will be 14203.

When $n$ is even, as we vary $k$ , the smallest integer $X$ will be $\boxed {4492}$ .

$\sqrt{2017}$ is not an integer, so it cannot be $X^2$ .

$\sqrt{20170\cdots0}<X<\sqrt{20180\cdots0}$ , for the same number of trailing 0's:

$142<\sqrt{20,170}<X<\sqrt{20,180}<143$ , not possible.

$449<\sqrt{201,700}<X<\sqrt{201,800}<450$ , not possible.

$1420<\sqrt{2,017,000}<X<\sqrt{2,018,000}<1421$ , not possible.

$\sqrt{20,170,000}<4492\leq X\leq4492<\sqrt{20,180,000}$ , possible for $X=4492$ .

The square root of 2017 is 44.9... which is not an integer.

The next possible answer is the smallest positive integer of the square roots of 2017n (5 digits) where n is 0-9. We could try all 10 values in ascending order and stop if we find a positive integer, but there is a faster approach. In order for X^2 to be within 2017n (let's call that the "solution range"), X must be must be between the square roots of 20170 and 20179, i.e. there must be a positive integer between the square roots of 20170 and 20179. In order for this to be true, the integer portion of the square root of 20179 must be strictly larger than the integer portion of the square root of 20170. If not, we will keep adding a digit until there is a positive integer between the square roots of the smallest and largest numbers in the solution range .

For 2017n:

• square root of 20170 = 142.02...
• square root of 20179 = 142.05...

There is no positive integer between 142.02... and 142.05...

For 2107nn:

• square root of 201700 = 449.11...
• square root of 201799 = 449.22...

There is no positive integer between 449.11... and 449.22...

For 2107nnn:

• square root of 2017000 = 1420.21...
• square root of 2017999 = 1420.56...

There is no positive integer between 1420.21... and 1420.56...

For 2107nnnn:

• square root of 20170000 = 4491.10...
• square root of 20179999 = 4492.21...

There is one positive integer between 4491.10... and 4492.21... and it is 4492 .

If there were more than one positive integer in the range, we would just choose the smaller of the two.

Note that 44^2=1936,45^2=2025.So we must go to 440.(Since we can't get into fractions,hence we increase 1 place of decimal).440^2=193600,449^2=201601(2016...so close!!),but 450^2=202500,we now move to 4491^2=20179081,4492^2=20178064=2017....& job is done. :)

long x = 1;

long y;

void setup() {

Serial.begin(9600);

}

void loop() {

y = pow(x, 2);

Serial.println(y);

while (y >= 10000) {

y = y / 10;

}

if (y == 2017) {

Serial.print("The value of x is: ");

Serial.print(x);

while (1) {

}

}

x++;

}

The code above gives the value of x in a couple of seconds.