A is the smallest positive integer which when divided by 9 and 12 leaves remainder 8. B is the smallest positive integer which when divided by 9 and 12 leaves remainder 5. Which one of the following is the value of A – B ?
[amp_mcq option1=”3″ option2=”2″ option3=”1″ option4=”0″ correct=”option1″]
This means A $\equiv$ 8 (mod 9) and A $\equiv$ 8 (mod 12).
Equivalently, A – 8 is divisible by both 9 and 12. Therefore, A – 8 must be a multiple of the LCM of 9 and 12.
The prime factorization of 9 is $3^2$. The prime factorization of 12 is $2^2 \times 3$.
LCM(9, 12) = $2^2 \times 3^2 = 4 \times 9 = 36$.
So, A – 8 must be a multiple of 36. A – 8 = 36k for some integer k. A = 36k + 8.
We are looking for the smallest positive integer A.
If k=0, A = 36(0) + 8 = 8. Check: $8 \div 9$ gives remainder 8, $8 \div 12$ gives remainder 8. 8 is a positive integer.
So, A = 8.
B is the smallest positive integer which when divided by 9 and 12 leaves remainder 5.
This means B $\equiv$ 5 (mod 9) and B $\equiv$ 5 (mod 12).
Equivalently, B – 5 is divisible by both 9 and 12. Therefore, B – 5 must be a multiple of LCM(9, 12) = 36.
So, B – 5 = 36m for some integer m. B = 36m + 5.
We are looking for the smallest positive integer B.
If m=0, B = 36(0) + 5 = 5. Check: $5 \div 9$ gives remainder 5, $5 \div 12$ gives remainder 5. 5 is a positive integer.
So, B = 5.
Finally, we need to find the value of A – B:
A – B = 8 – 5 = 3.