Question That Changes with Time
As an example, for the past year - $2016\,$ - one solution is $2016=2007+2+0+0+7.\,$ Now, on to the present year - $2017.$
Obviously, we are looking for a four;three;four;five digit number, say, $N.\,$ So, assume $N=abcd=1000a+100b+10c+d.\,$ The sum in question is, say,
Clearly, $a\,$ could possibly be either $1\,$ or $2.\,$ Let's first check $a=1.\,$ The equation then reduces to $100b+10c+2d+b+c=1016.\,$ Taking the greatest three digit number we overshoot: $999+27=1026.\,$ Letting $d\,$ change,we find that $994+22=1016,\,$ giving us one solution 1994;$1993$;$1994$;$1995$;$1996$.
Perhaps, it could be possible to also reduce $c\,$ by taking $c=8\,$ and trying to solve
The largest of these, 989;$987$;$988$;$989$, gives $989+26=1015\,$ which is a little short. Thus, with $a=1\,$ we have just one solution;one solution;two solutions
Consider now the case of $a=2.\,$ The equation becomes $100b+10c+2d+b+c=15,\,$ which immediately gives b=0;$b=0$;$b=1$;$b=2$, simplifying the equation to $10c+2d+c=15.\,$ Clearly, $c\,$ can only be 1;$0$;$1$;$2$ which simplifies the equation even further: $2d=4\,$ such that $d=2\,$ and giving the answer as 2012;$2011$;$2012$;$2013$
Thus we have two answers;one answer;two answers;three answers: 1994,2012;1993,2011;1994,2011;1994,2012
We may return to the question next year when the time comes.
I found the question for the year 2010 at the Crux Mathematicorum.
Copyright © 1996-2017 Alexander Bogomolny