Question

Consider an arithmetical progression of positive integers . Prove that one can find infinitely many terms the sum of whose decimal digits is the same .

Answer 1

Let $a$ be the base and $b$ be the step of an arithmetic progression S, so that S has the form $(a,a+b,a+2b,...)$
Let the decimal expansion of a be $a_k...a_2{a}_1$ and the decimal expansion of b be $b_l...b_2{b}_1$. Then the elements $a+{10}^{k}b,a+{10}^{k+1}b,a+{10}^{k+2}b$, ... of S have expansions $bl...b_2{b}_1{a}_k...a_2{a}_1,b_l...b_2{b}_{1}0a_k...a_2{a}_1,b_l...b_2{b}_{1}00a_k...a_2{a}_1$, ..., whose sums of digits are all equal to ${\sum }_{i=1}^k{a}_i+{\sum }_{j=1}^l{b}_j$