What is the smallest positive integer n such that the rightmost three digits of n^2 and (n+2)^2 are the same?
+373 Hey there, Guest!
So, let's solve your problem:
Step 1: Simplify both sides of the equation.
\(n^2=n^2+4n+4\)
Step 2: Subtract n^2 from both sides.
\(n^2−n^2=n^2+4n+4−n^2\)
\(0=4n+4\)
Step 3: Flip the equation.
\(4n+4=0\)
Step 4: Subtract 4 from both sides.
\(4n+4−4=0−4\)
\(4n=−4\)
Step 5: Divide both sides by 4.
\(\frac{4n}{4}=\frac{-4}{4}\)
Therefore, n=-1.
Hope this helped! :)
( ﾟдﾟ)つ Bye
+876 Nice job, TaliaArticula! Not so good at NT myself. Awesome and detailed solution!
