+817 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 $b^2?$
+121 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 $899$.
+129771 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
