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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?

jvdhooft
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akira
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1 Answers1

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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.

JC12
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  • Thanks JC12! Gosh so simple, I over complicated this trying to solve this with algebra.... – akira Apr 20 '20 at 07:29
  • The lowest common multiple of 6 and 8 is 24. So we need the minimum multiple of 24 that's over 50, which is 72. – IanF1 Apr 20 '20 at 14:17