Which of the following numbers is divisible by $3$?

### Concept & Logic The problem tests the standard divisibility rule for $3$. A number is perfectly divisible by $3$ if and only if the sum of all its individual digits is a multiple of $3$. ### Step-by-Step Solution **Calculation:** We evaluate the given options by summing their digits. Test Option A: $541326$ Sum of digits $= 5 + 4 + 1 + 3 + 2 + 6 = 21$ Since $21$ is divisible by $3$ ($21 / 3 = 7$), the number $541326$ is divisible by $3$. Test Option B: $5967013$ Sum of digits $= 5 + 9 + 6 + 7 + 0 + 1 + 3 = 31$ Since $31$ is not divisible by $3$, the number is not divisible by $3$. (We can stop testing options as we have found the correct one). ### Exam Strategy & Shortcut Instead of adding all digits, use the ''digit cancellation'' method. Quickly scan the number and cross out any digits or pairs of digits that intuitively add up to $3, 6,$ or $9$. For $541326$: cross out $3$, cross out $6$, cross out $5+4$ ($9$), and cross out $1+2$ ($3$). Since all digits are crossed out with no remainder, it is divisible by $3$. This takes two seconds per option. ### Common Pitfall Attempting to actually divide the entire large number by $3$ using long division. This wastes precious exam time and increases the likelihood of a calculation error. ### Final Answer Therefore, the correct answer is **541326**.