I had a slight question on this explanation. You had stated that n can be "
6, 15, 24, ..." and m can be "
4, 16, 28.....".
Would it be technically not possible for n to be 6 or m to be 4 since we need a quotient of at least 1 for the problem to work? For example, a bag of n peanuts can be divided into 9 smaller bags with 6 peanuts left over. Okay, there, at least 1 smaller bag HAS to be an outcome, so n =6 cannot be a possibility as that leaves the quotient as 0 [in other words, 0 'smaller bags of peanuts'].
Similar line of reasoning for why m cannot be 4.
Showmeyaa
Bunuel
A bag of n peanuts can be divided into 9 smaller bags with 6 peanuts left over. Another bag of m peanuts can be divided into 12 smaller bags with 4 peanuts left over. Which of the following is the remainder when nm is divided by 18?
A. 3
B. 5
C. 6
D. 10
E. 122
Given:
\(\frac{n}{9}, R = 6\)
So, n can be 6, 15, 24......
Also given:
\(\frac{m}{12}, R = 4\)
m can be 4, 16, 28.....
Therefore, mn can be 24, 96...(you can pick any combination)
\(\frac{mn}{18}\)
\(\frac{24}{18}, R = 6\) - (C), IMO!