A box contains some red and some yellow balls. If one red ball is removed, one seventh of the remaining balls would be red; if one yellow ball is removed, one-sixth of the remaining balls would be red. If n denotes the total number of balls in the box, then the sum of the digits of n is:
I'm pretty sure that n=43. But the answer given is that sum of digits is 6. Can you please check my answer.
The answer given is wrong, in fact let $T$ the total number of balls and let $R$ the number of red ones. So, we have: $$\left\{\begin{matrix} R-1=\frac{1}{7}(T-1) \\ R=\frac{1}{6}(T-1) \end{matrix}\right.$$ Substracting the first to the second, we have: $$1=(T-1)\cdot\frac{1}{42}\leftrightarrow T=43$$ So, the sum of digits of $T$ is $7\neq6$.