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 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: 3,372 Thanks: 233 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 , re-arrange , $9x = 12 - 19k$ $\ \ \ \ \ \$ <--- (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 Thanks from greg1313
 January 6th, 2017, 04:24 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,045 Thanks: 1618 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. Thanks from greg1313 and agentredlum
January 6th, 2017, 09:14 AM   #4
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Quote:
 Originally Posted by agentredlum I can offer an alternative solution. We are looking for positive integers $x$ , $y$ which satisfy
it looks like what you've done is apply the Euclidean algorithm in disguise.

 January 6th, 2017, 08:45 PM #5 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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.

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