Weekly brain teaser | A gang of 17 thieves steals a bag of gold coins

fight it out Good luck

Every week we feature a Brain Teaser or a thought-provoking question.

If you have one that you would like to share with fellow engineers please email it to info@engineeringdaily.net.

So, here is this week’s,

A gang of 17 thieves steals a bag of gold coins. In trying to share the coins equally, there are three coins remaining. In the ensuing fight over these three coins, one of the gang members is killed. In the next attempt to equally distribute the coins, there are 10 coins remaining. Again the gang fights, and another member dies. The third attempt is successful.

What is the smallest number of coins stolen?

The correct answer will be posted a week following this posting.

Good luck.


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    1. Of course, that 16th guy was certainly not minding his comparative values, deciding to fight and die rather than simply pony up 5 of his 245 coins to ensure his own safety.

      If someone is going to die over that last batch of 10 leftovers, and once it has been established that the others are willing to turn on and kill off any one of the others over only 3 coins, it would have been much better for him to have determined which of the others was the least likely to assist him in future endeavors and ‘volunteer’ that individual as the scapegoat, netting him 486 gold (out back) for his initial investment of 5, and ensuring it is not the other 15 that split his own take…and head…hmmm

    2. Even with the admin answer I could not see a mathematical substitution method to answer the question using the 3 equations available. So I cheated, using an EXCEL data base with three columns, one for each equation, with the values then placed in one column in ascending order will yield 3 consecutive 3930 values. To “spot” it, a simple IF-AND equation like =IF((H2-H1=0)*AND(H3-H1=0),”yes”,””), identifies where the three equations will all yield the same value. Knowing the 3930 answer we can simply solve for the admin’s “a, b, & c” values. Not the way Einstein would solve it, but it works.

    3. Dont’ forget the last part that the total is equally divisible among 15 thieves. The 272 is strictly a multiplier between 16 and 17. I found the solution by finding the least common multiple of 15x, 16x+10, and 17x+3. This way the remainers (10 and 3) are accounted for according to the the number of thieves.

    4. 3930. The number of coins. X = 17a + 3 = 16b + 10 = 15c. By inspection, the first two criteria are met when a=b=7 and x=122. They are subsequently met at increments of x of 16×17 or 272. The first number in this series divisible by 15 is 3930.

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