Brain Teaser - A gang of 17 thieves steals a bag of gold coins

Professional Engineer & PE Exam Forum

Help Support Professional Engineer & PE Exam Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.
If we're talking pennies or dimes, I don't think it's that big. But would seventeen pirates be willing to fight and kill over that amount? I doubt it.
See .. that's just it benbo ... it they were ninjas ... we wouldn't even be having this discussion!!! :ph34r:

17 thieves were required to steal the bag of coins because they were unionized, the number of coins is irrelevent due to the useless expendenture of union dues. The real question should have been, "How long does it take to have 17 thieves steal a bag of coins, disregarding seniority".
Zing !!!

The theft was run by gov contractors?


Zing +1 !!!!

JR

 
This thread has gone long enough without an answer. I got

3930

, but it seems high to be the "minimum". Anybody confirm/deny?

 
This thread has gone long enough without an answer. I got
3930
, but it seems high to be the "minimum". Anybody confirm/deny?
Confirm, at least that's what I got.

 
Last edited by a moderator:
Wait, how did these thieves get the gold coins? Who has these gold coins?

 
correct.... but why did you decide to deviate from the theme of the thread and actually do calculations?!?

 
correct.... but why did you decide to deviate from the theme of the thread and actually do calculations?!?
I am expected to deviate, I work for the government, and yes, I am here to help.

 
SapperPE said:
I tried to do it in my head but then realized it would take me entirely too long to work that out at midnight, so I let it go and read the rest of the thread.
Much better plan!

 
Warning: Serious answer follows... if you were hoping for more "sack" responses, you'll be disappointed.

This is a classic "Chinese Remainder Theorem"...

The problem requires some n where:

n ≡ 3 (mod 17)

n ≡ 10 (mod 16)

n ≡ 0 (mod 15)

If you're not familiar with modulo notation, the first one says that some number n and 3 differ by 17... or in words we're taught in third grade, n divided by 17 has a remainder of 3.

OK... so the approach is to find some x, y, and z such that n = x + y + z with the following constraints:

x ≡ 3 (mod 17), which satisfies the first requirement and y and z are multiples of 17 (so adding them doesn't change the remainder)

y ≡ 10 (mod 16), which satisfies the second requirement and x and z are multiples of 16 (so adding them doesn't change the remainder)

z ≡ 0 (mod 15), which satisfies the third requirement and x and y are multiples of 15 (so adding them doesn't change the remainder)

And so this really means:

x ≡ 3 (mod 17) and x is a multiple of (15)(16) = 240

y ≡ 10 (mod 16) and y is a multiple of (15)(17) = 255

z ≡ 0 (mod 15)... and this is already satisfied by x and y being a multiple of 15, so we can forget about it!

OK then... let's find an x that works:

240 / 17 = 14 R2

480 / 17 = 28 R4

720 / 17 = 42 R6

960 / 17 = 56 R8

1200 / 17 = 70 R10

1440 / 17 = 84 R12

1680 / 17 = 98 R14

1920 / 17 = 112 R16

2160 / 17 = 127 R1 (126 R18)

2400 / 17 = 141 R3 (140 R20) <---- this is the one!

And then let's find a y that works:

255 / 16 = 51 R15

510 / 16 = 31 R14

765 / 16 = 47 R13

1020 / 16 = 63 R12

1275 / 16 = 79 R11

1530 / 16 = 95 R10 <---- this is the one!

So... n = 2400 + 1530 = 3930

 
Last edited by a moderator:
....or in words we're taught in third grade, n divided by 17
for the West Virginians that's "or in words we're taught in 8th grade 17 gozzinta n...."

 
Last edited by a moderator:

Latest posts

Back
Top