The prime numbers dividing 143 and leaving a remainder of 3 in each case are

### Concept & Strategy This problem combines divisibility with remainder logic. If a number $N$, when divided by a divisor $d$, leaves a remainder $R$, then the number $(N - R)$ must be perfectly divisible by $d$. $$N \equiv R \pmod d \implies (N - R) \text{ is a multiple of } d$$ ### Step-by-Step Solution Given $N = 143$ and the remainder $R = 3$. Subtract the remainder from the main number to find the perfectly divisible value: $$143 - 3 = 140$$ The required prime numbers must be factors of $140$. Let's find the prime factorization of $140$: $$140 = 14 \times 10$$ $$140 = (2 \times 7) \times (2 \times 5)$$ $$140 = 2^2 \times 5 \times 7$$ The unique prime factors of $140$ are $2, 5$, and $7$. This means dividing $143$ by $2, 5$, or $7$ will leave a remainder of $3$ (provided the divisor is greater than the remainder, so $2$ is technically invalid as a divisor leaving a remainder of $3$, but $5$ and $7$ are valid). Now, let's look at the given options to find a pair of these valid prime factors: * (a) 2 and 11: 11 is not a factor of 140. * (b) 11 and 13: Neither are factors of 140. * (c) 3 and 7: 3 is not a factor of 140. * (d) 5 and 7: Both are prime factors of 140. ### Exam Strategy & Shortcut Instead of factoring, use the Options Elimination technique by testing the divisors directly on $143$. Test option (d) 5 and 7: * $143 \div 5 = 28$ with remainder $3$. (Works!) * $143 \div 7 \rightarrow 140$ is a multiple of $7$, so $143$ leaves remainder $3$. (Works!) This avoids algebraic setup and lands you the answer in seconds. ### Common Pitfall A common mistake is trying to divide $143$ directly by all options without mentally subtracting the $3$ first. Mentally checking if $140$ is a multiple of the options is vastly quicker than calculating remainders for $143$. ### Final Answer Therefore, the correct answer is 5 and 7.