New Year Puzzle

A four-digit number, \(\overline{abcd}\) (in base 10) can be expressed as \(\overline{abcd}=1000a+100b+10c+d\). If we change this formula to \(998a+102b+98c+a+1\), for \(1000<\overline{abcd}<9999\), how many times does \(\overline{abcd}=998a+102b+98c+d+1\)?

×

Problem Loading...

Note Loading...

Set Loading...