## gcd and divisibility

For discussing Olympiad Level Number Theory problems
kh ibrahim
Posts: 17
Joined: Mon May 09, 2016 11:18 am

### gcd and divisibility

If a multiple of 864 is taken which is a positive integer,then what is the probability of that number to be divisible by 1944

nahin munkar
Posts: 81
Joined: Mon Aug 17, 2015 6:51 pm
Location: banasree,dhaka

### Re: gcd and divisibility

The ans is $\dfrac{1}{ 9}$.
Last edited by nahin munkar on Mon Aug 22, 2016 1:57 pm, edited 1 time in total.
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss

kh ibrahim
Posts: 17
Joined: Mon May 09, 2016 11:18 am

### Re: gcd and divisibility

can u describe the probability part a bit more clearly?

asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm

### Re: gcd and divisibility

Take $864k$ as a multiple of $864$. Now if $1944$ divides $864k$, we must have $\dfrac{864k}{1944}\in \mathbb{N}$. Which implies $9\mid k$. The probability of happening this is $\dfrac{1}{9}$ when we choose $k$ randomly.

kh ibrahim
Posts: 17
Joined: Mon May 09, 2016 11:18 am

### Re: gcd and divisibility

I got that divisibility part that 9 is the factor of k but I I didn't get the probability part.if u could interpret how to determine the probability It would be percieved...thanx

nahin munkar
Posts: 81
Joined: Mon Aug 17, 2015 6:51 pm
Location: banasree,dhaka

### Re: gcd and divisibility

kh ibrahim wrote:I got that divisibility part that 9 is the factor of k but I I didn't get the probability part.if u could interpret how to determine the probability It would be percieved...thanx
OK . If $9|k$, then $k$ is divided by $9$ only when it is a multiple of $9$. Then u will get $k$ as a multiple of $9$ after each $9$ numbers . Observe $9,18,27,36..$...etc are multiple of $9$ .Look they r returning back as each $9th$ number ($8$ number after after). If u make use of probability formula then u get that $k$ returns as a multiple of $9$ after each $9$ numbers. So, u'll get 1 number from any consequent $9$ numbers. Mainly, the probability version tells us u will get 1 number k as multiple of $9$ if u choose $9$ numbers.So, the probability is $\dfrac{1}{9}$ as follows.
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss