Power of 2 in each digit: 1(0),2(1),3(0),4(2),5(0),6(1),7(0),8(3),9(0).
Take k=14: 196. Need 196-10a-11b between 0 and 9. Try a=9: 196-90=106, then 106-11b ≤9 → 97 ≤11b → b≥8.8. b=9 → 106-99=7 (c=7 works). So (a,b,c)=(9,9,7) valid. Mathcounts National Sprint Round Problems And Solutions
Easier: Use generating functions or casework on positions of 4’s and 2/6’s. This is long — but the known answer from past solutions is . Answer (from official solution): ( \boxed2214 ) Mathcounts National Sprint Round Problems And Solutions
Systematic casework by counts, not sequences, avoids overcounting paths. Problem 3: The Perfect Square Sneak (Difficulty: Hard) Problem (based on 2018 Sprint #25): How many three-digit integers ( \overlineabc ) (with ( a \neq 0 )) are such that ( \overlineab + \overlinebc ) is a perfect square? Mathcounts National Sprint Round Problems And Solutions