What is the third term in a sequence of numbers that leave remainders of $1$, $2$ and $3$ when divided by $2$, $3$ and $4$ respectively?

### Concept & Logic The remainders are $1, 2, 3$ for divisors $2, 3, 4$. Notice the difference between Divisor and Remainder is constant: $2-1 = 1$, $3-2 = 1$, $4-3 = 1$. The general number is $N = \text{LCM}(2, 3, 4)k - 1$. ### Step-by-Step Solution * **Find LCM:** $\text{LCM}(2, 3, 4) = 12$. * **Formulate sequence:** $N = 12k - 1$. * **Find terms:** For $k=1$, $N = 12(1) - 1 = 11$ (First term). For $k=2$, $N = 12(2) - 1 = 23$ (Second term). For $k=3$, $N = 12(3) - 1 = 35$ (Third term). ### Exam Strategy & Shortcut Once you find the first term ($11$) and the LCM ($12$), the subsequent terms in the sequence are just an arithmetic progression: $11, 11+12=23, 23+12=35$. ### Common Pitfall Confusing the sequence with a simple $+1$ pattern. Always calculate the LCM of the divisors to determine the common difference for the sequence. ### Final Answer **Therefore, the correct answer is 35.**