Try this beautiful problem from the PRMO II, 2019 based on Sum of Digits base 10.
Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m)=20 and s(33m)=120, what is the value of s(3m)?
But try the problem first...
Answer: is 60.
PRMO II, 2019, Question 7
Elementary Algebra by Hall and Knight
taking sum of digit base 10 to (mod 9)
and s(ab)=s(a).s(b)(mod 9)
[ let x congruent r mod n, y congruent to s mod n,
\(0 \leq r,s \leq n-1\),
x=in+r, y=jn+s, i,j are integers
xy=(in+r)(jn+s)=ij\(n^2\)+(is+jr)n+rs congruent to rs mod n
so, xy mod n =(x mod n)(y mod n) ]
s(33m)=120=\(s(11) \times s(3m)\)
or, 120=\(2 \times s(3m)\) [ since s(11)=2(mod 9)]