BECE Mathematics 2010 Paper

Concept:In base $5$, the only valid digits are $0,1,2,3,4$.After a digit reaches $4$, the next number involves a carry, just like $9$ followed by $10$ in base $10$.Explanation:Write the given sequence in base $5$: $32,33,34,\ldots,42$.Counting forward in base $5$:$32_{\text{five}} \rightarrow 33_{\text{five}} \rightarrow 34_{\text{five}} \rightarrow 40_{\text{five}} \rightarrow 41_{\text{five}} \rightarrow 42_{\text{five}}$.This is because after $34_{\text{five}}$, the units digit $4+1$ equals $5$, but $5$ is not allowed as a digit in base $5$.So we write $0$ and carry $1$ to the tens place: $34_{\text{five}} + 1 = 40_{\text{five}}$.Then continue: $40_{\text{five}} \rightarrow 41_{\text{five}} \rightarrow 42_{\text{five}}$.Therefore, the missing numbers in the sequence are $40_{\text{five}}$ and $41_{\text{five}}$.Answer:Option C: $40,41$