January 5th, 2017, 09:03 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, 04:47 AM  #2 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,140 Thanks: 157 
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, 05:24 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 16,369 Thanks: 1172 
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, 10:14 AM  #4 
Senior Member Joined: Sep 2015 From: CA Posts: 753 Thanks: 399  
January 6th, 2017, 09:45 PM  #5 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,140 Thanks: 157 
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 07:28 PM. 
January 10th, 2017, 07:09 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.


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