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help I have question about squares

 

Let $a$ and $b$ be non-zero integers.  If both $a^2$ and $b^2$ have at most three digits, then what is the greatest possible difference between $a^2$ and $b^2?$

 Aug 14, 2023
 #1
avatar+121 
+1

Let's analyze the problem step by step:

We are given that both $a^2$ and $b^2$ have at most three digits. This means that the largest possible value for a three-digit number is $999$, and the smallest possible value for a three-digit number is $100$. So, we have the range for $a^2$ and $b^2$ as $100 \leq a^2, b^2 \leq 999$.

To maximize the difference between $a^2$ and $b^2$, we should maximize $a^2$ and minimize $b^2$. The largest possible value for $a^2$ is $999$ (since it's at the upper limit of the range), and the smallest possible value for $b^2$ is $100$ (since it's at the lower limit of the range).

So, the greatest possible difference between $a^2$ and $b^2$ is:

$$\text{Difference} = a^2 - b^2 = 999 - 100 = 899.$$

Therefore, the greatest possible difference between $a^2$ and $b^2$ is $899$.

 Aug 14, 2023
 #2
avatar+129197 
+1

Note that both a^2 and  b^2   have AT MOST three digits

But this means that a^2  could have three digits but b^2  could have only one  digit

 

Since both a and  b are non-zero integers the largest that a could be = 31 so that 31^2   = 961

And the smallest that b could be = 1 so that 1^2 = 1

 

So  the greatest difference   =

 

a^2 - b^2  =  961  - 1 =   960

 

 

cool cool cool

 Aug 14, 2023
edited by CPhill  Aug 14, 2023

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