the number of perfect square natural number between 100 and 10000 is equivalent with $100<{ x }^{ 2 }<10000$

$\sqrt { 100 } <\sqrt { { x }^{ 2 } } <\sqrt { 10000 }$

$10<x<100$

thus, there is 89 natural number between 10 and 100

Finding number of Perfect squares between 100 and 10000 is equivalent to finding number of numbers between 10 and 100, because 10^2 = 100 and 100^2 = 10000. All the numbers will lie between them.

Thus the answer is 89

10² is 100, just one smaller than 101, so we start counting from 11.

100² is 10000, which is one larger than 9999, so we stop counting at 99.

From 11 to 99 there are 89 numbers, so the answer is 89 .

According to given condition 100<x^2<10000,,,,,,, taking square root we get 10<x<100........ There are total 89 natural numbers between 10 to 100... So ANSWER IS ""89"".

Write a solution. basically this is a sum that can be solved by using the range of square since 10^2=100 and 100^2=10000 then all the numbers between them will have squares in this range thus there are 89 numbers between them thus the answer is 89

this is my solution program in c:

include<stdio.h>

include<math.h>

void main() { int j,k,l=0; for(j=101;j<10000;j++) { for(k=1;k<j;k++) { if(j==k*k) l++; } } printf("%d",l); }

121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500.......... In these, 101~400: 10 ps 401~900: 10ps 901~1600: 10ps 1601~2500: 10ps 2501~3600: 10ps 3601~4900: 10ps 4901~6400: 10ps 6401~8100: 10ps 8101~10000: 10ps So 90 ps upper 101 below 10000. But 101~9999, so 10000 exceeds. So 90-1=89 perfect squares(ps meant this.)

A pattern is that the sequence of "the difference between consecutive perfect squares (e.g. 1, 4, 9, 16)" is a sequence of consecutive odd numbers (e.g. 4-1= $\textbf{3}$ , 9-4= $\textbf{5}$ , 16-9= $\textbf{7}$ .)"

Another way to count the number of perfect squares is by counting the number of consecutive odd differences between consecutive perfect squares in [101, 9999]. $11^2-10^2$ =121-100=21. $100^2-99^2$ =10000-9801=199. Between 101 and 9999, the sequence of perfect squares is as follows: (100+21), (121+23), ..., y+195, x+197. (9801+199) is not included. From 21 to 197 there are (197-21)/2= 88 increments by two. 88 consecutive increments imply 89 consecutive odd integer differences. By the pattern of consecutive odd integers as differences between consecutive perfect squares, we can count 89 perfect squares.

Python one-liner:

print(len([n for n in range(101, 10000) if (n**0.5).is_integer()]))

sqrt(100)=10 and sqrt(10000)=100

we have to find perfect squares between 100 and 10000 (not inclusive) which will be

(100-10)-1=89

10 10=100 it cant be counted. so start with 11 11 i.e 121 and at last 100*100=10000 it also can not be counted so total no =99-10 i.e 89. thats shortcut. 10 and 100 should not be counted so count 11 to 99 both inclusive.

100 is 10^2 and 10000 is 100^2 so all the numbers between 10 and 100 have squares between 100 and 1000

there are 89 no. s between 10 and 100 so there r 89 perfect squares between 100 and 10000

100=10^2 and 10000=100^2 So, If we include 10 and 100 the count of numbers between 10 and 100 (both inclusive is 91) and as the question says squares of 10 and 100 are not to be included thus (91-2=89).

Here is my JavaScript program (you can run it directly in your browser, for Chrome press CTRL + Shift + J):

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var counter = 0;

for (var x = 101; x < 10000; x++) {
        var y  =  Math.sqrt(x);

        if (y % 1 === 0)
                counter++;
}

alert(counter); 

10^2,(10+1)^2,(10+2)^2,(10+3)^2,..........,(10+89)^2,(10+90)^2. Excluding 10^2 and (10+90)^2 89 perfect squares are there

The numbers including 10(root of 100) & 100(root of 10000) are 91. Therefore the numbers between will be 91-2=89. Thus the perfect squares between 100 & 10000 are 89.

So, I read the problem quickly and then try to solve it while biking to work, but I misremembered the question as being... How many perfect squares are there between 100 and 1000? I think this might be a more interesting question, since the square root of 1000 is not a whole number (31^2=961 & 32^2=1024). This requires you to figure out the square root of 100 (and maybe without a calculator?) So, if 10^2=100 and 31^2=961, then how many perfect squares are between 10^2 and 31^2? But that wasn't the real problem...

Which brings me to the same issue I had with the # of perfect squares between 100 and 10000? What does it mean to be "between"? Are 10^2 and 100^2 not included because they are the outside boundaries? Or should they be included because they equal 100 and 10000, respectively? In other words, do we mean exclusive or inclusive of the boundaries?

Exclusive of the boundaries => 89 Inclusive of the boundaries => 91

It is not clear from the question if the boundaries are inclusive or not. Obviously the answer can be extrapolated from the 4 options, but this seems cheating to me.

the quetion is simply asking about the natural numbers between 10 and 100 if we see clearly! :P