Select Page

# Understand the problem

A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?

##### Source of the problem
American Mathematics Competition

Number Theory

7/10

##### Suggested Book

Elementary Number Theory by David M. Burton

# Start with hints

Well, just give the problem a good read. Probably, with a little bit of thought, you can even get this done without a hint !

We could start this the traditional way, be assuming the number of coins to be x.  Now, ask yourself after the k’th pirate has taken his share, what is the remanant number of coins ? This is  ( 12-k / 12 ) of what was originally there. [ Why ? Because each pirate takes k/12 of the coins, remember ? ]  Now, could you try taking things up from here…by yourself ?

Let’s understand the next thing the problem is trying to focus on. “Each pirate receives a whole number of coins” Now, this should actually help us conclude   x. ( (11.10.9.8.7.6.5.4.3.2.1) / 12 ) is supposed to be an integer. Since this actually implies divisibility.   Cancellation of terms leads us to : x. ( (11.5.1.7.1.5.1.1.1.1) / ( 12.6.2.12.2.12.3.4.6.12 ) )  Can you try and approach the solution by yourself now ?

Now, this tells us the intuition of the problem. We make sure that the quotient should be an integer ! Also, recall that the 12’th pirate definitely takes the entirety of what is left, practically unity since it is exactly divisible.

So basically, we just realized that the denominator is entirely multiplied out…cancelled !  And since we know that the denominator cancels out, the number of gold coins received by the 12th pirate is just going to be the product of the numerators !!! That evaluates to : 11.5.7.5 = 1925 And that completes our solution !

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## Right-angled shaped field | AMC 10A, 2018 | Problem No 23

Try this beautiful Problem on triangle from AMC 10A, 2018. Problem-23. You may use sequential hints to solve the problem.

## Area of region | AMC 10B, 2016| Problem No 21

Try this beautiful Problem on Geometry on Circle from AMC 10B, 2016. Problem-20. You may use sequential hints to solve the problem.

## Coin Toss Problem | AMC 10A, 2017| Problem No 18

Try this beautiful Problem on Probability from AMC 10A, 2017. Problem-18, You may use sequential hints to solve the problem.

## GCF & Rectangle | AMC 10A, 2016| Problem No 19

Try this beautiful Problem on Geometry on Rectangle from AMC 10A, 2010. Problem-19. You may use sequential hints to solve the problem.

## Fly trapped inside cubical box | AMC 10A, 2010| Problem No 20

Try this beautiful Problem on Geometry on cube from AMC 10A, 2010. Problem-20. You may use sequential hints to solve the problem.

## Measure of angle | AMC 10A, 2019| Problem No 13

Try this beautiful Problem on Geometry from AMC 10A, 2019.Problem-13. You may use sequential hints to solve the problem.

## Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Try this beautiful Problem on Geometry from PRMO -2018.You may use sequential hints to solve the problem.

## Recursion Problem | AMC 10A, 2019| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-15, You may use sequential hints to solve the problem.

## Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-24, You may use sequential hints to solve the problem.

## Set of Fractions | AMC 10A, 2015| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2015. Problem-15. You may use sequential hints to solve the problem.