January 5th, 2017, 08:03 PM  #1 
Newbie Joined: Dec 2016 From: Dhaka, Bangladesh Posts: 9 Thanks: 0  multiple of a number
Divide 1000 into two portions in such a way that one part or one portion will be multiple of 47 and other number will be multiple of 19. I cannot understand how this problem can be solved. The source from which I have collected this problem has given the following solution of this problem. But still now the solution seems very hard and awkward to be understood. I would be grateful if anyone of this group/forum will show me alternative ways to solve the problem stated above or make me understand the solution I have given below. 47+19=66×15=990 Rest 10+47=57 is divisible by 19. ∴47×14=658 and 19×(15+3)=342. ∴ 1000 can be divided into 658 and 342. 
January 6th, 2017, 03:47 AM  #2 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,347 Thanks: 227 
I can offer an alternative solution. We are looking for positive integers $x$ , $y$ which satisfy $47 x + 19y = 1000 $ $ 19y = 1000  47x $ $ y = (\frac{1000}{19})  (\frac{47x}{19}) $ $y = (52 + \frac{12}{19} )  (2x + \frac{9x}{19})$ Now ... throw away $52$ and $2x$ because they are integers already Forget $y$ for now , we need $\frac{12  9x}{19}$ to be an integer , soooo.... $\frac{12  9x}{19} = k $ for integer k , rearrange , $9x = 12  19k$ $ \ \ \ \ \ \ $ < (we will come back to this later) *** $x = (\frac{12}{9}) (\frac{19k}{9})$ $x =(1 + \frac{3}{9} ) (2k + \frac{k}{9})$ guess what? Throw away $1$ and $2k$ because they are integers already and forget about $x$ for now we need $\frac{3 k}{9}$ to be an integer , now it's easy to choose $k = 3$ but that will make $x$ negative. Choose $k = 6$ and substitute into the equation I labeled *** above $9x = 12  19(6)$ $9x = 126$ $x= 14$ next $47(14) + 19y= 1000$ $y = \frac{1000  658}{19}$ $y = 18 $ Not the fastest way to get the answer but I hope it's logical 
January 6th, 2017, 04:24 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 17,734 Thanks: 1360 
The given solution starts with 990 = 15(47 + 19) = 15*47 + 15*19. Conveniently, 10 + 47 = 57 = 3*19, so 10 = 47 + 3*19. Adding gives 1000 = 14*47 + 18*19 = 658 + 342. 
January 6th, 2017, 09:14 AM  #4 
Senior Member Joined: Sep 2015 From: CA Posts: 1,302 Thanks: 666  
January 6th, 2017, 08:45 PM  #5 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,347 Thanks: 227 
We are looking for positive integer solutions to $47x + 19y = 1000$ The Euclidean Algorithm would find $ \ \ \ $ $gcd(47 , 19) = 1$ $ \ \ \ $ then reversing the steps would give a way to express that $gcd$ , namely $1$ , as an integer combination of $47$ and $19$. We would get a solution pair $(a , b)$ to $47a + 19b = 1$ $ \ \ \ \ \ \ $ < (Obviously one of $a$ , $b$ is going to be negative) Then we would need to multiply through by $1000$ and do considerably more work to reduce the answer to the one they want. It is interesting to note that if you get a solution that is close to the one they want you can easily get to the solution they want. For example: $47(5)+ 19(65) = 1000$ is true but doesn't satisfy the requirement that both numbers be positive but, $5 +19 = 14$ and $65 47 = 18$ $ \ \ \ \ \ $ < (note the use of $19$ and $47$) $x = 14$ and $y = 18$ satisfy the requirement So ... did I use the Euclidean Algorithm in disguise? I would say no. Last edited by skipjack; January 7th, 2017 at 06:28 PM. 
January 10th, 2017, 06:09 AM  #6 
Newbie Joined: Dec 2016 From: Dhaka, Bangladesh Posts: 9 Thanks: 0 
I was really unlucky since I missed this beautiful discussion due to disconnection of internet. Really this explanation was very much helpful.


Tags 
multiple, number 
Search tags for this page 
divide 1000 into two portions in such a way that one part will be multiple of 47 & other number will
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Multiple  Django  Math  7  January 24th, 2016 01:29 PM 
natural number multiple of another number if its digit sum equal to that number  Shen  Elementary Math  2  June 5th, 2014 07:50 AM 
Every natural number has multiple with digits 7 or 0  international  Number Theory  24  July 11th, 2013 10:43 PM 
probability of multiple events with multiple trials  prwells32  Advanced Statistics  4  January 30th, 2010 02:42 PM 
multiple of 11  4011  Number Theory  7  May 29th, 2007 10:29 AM 