Question

the student noticed that if he writes a multiplication sign between the second and third digit of number 2020 tje product 20.20 gives a perfect square . how many numbers between 2010 and 2099 (without 2020 ) have the same property​

Answer 1

Answer:

Step-by-step explanation:

We need 20*d to be perfect square, where  10 <= d <= 99

Let  20*d = (5c)^2 =>d=5c^2

=>10<=5c^2<=99 =>2<= c^2<=99/5 => c^2=4,9,10    

But  c^2=4 is already excluded

Answer 2

Let the number having the same property as 2020 be x

Now, x is a 4-digit no. between 2010 and 2099.

So, the condition we be,

$a\in ( 10,99)$ such that 20a is a perfect square.

Let 20a be $b^{2}$

then $b=\sqrt{20a}=2\sqrt{5a}$, which means $\sqrt{5a} \in Z$

For,$\sqrt{5a} \in Z$, $a=\frac{c^{2}}{5}$ or $a=5c^{2}$

Case 1: $a=\frac{c^{2}}{5}$

This means , $c=\sqrt{5a}, a\in (10,99)$

$\therefore a= \{20,45,80\}$

Case 2:$a=5c^{2}$

This means, $c=\sqrt{\frac{a}{5}}, a\in (10,99)$

$\therefore a= \{20,45,80\}$

So the other nos. are 2045 and 2080

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