Bob has no less than 50 $2 coins in a bag. After organizing them, he finds that there are no remainders whether it's 6 or 8 coins per row. What is the minimum total value of the coins?
Saw it on TV and the answer is $144. I can't seem to solve it algebraically. How can I solve this problem?
If it can be arranged in a row of 6 and 8 without a remained, then 6 and 8 is a factor of the number of coins. The lowest common multiple of 6 and 8 is 24, however 24 is lower than the minimum of 50 coins required to be in a bag. Through trial and error you can find that the smallest number with both 6 and 8 as a factor over 50 is 72. If you have 72 2 dollar coins, then the total value is $144, your answer.