The sum of two positive integers is $200$. If one is divided by $5$ and the other is divided by $9$, the remainder is $1$ each case. Find the numbers
I have $u+v=200$. But I can't get the second equation.
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If 5 divides $u$ then you can notice that you can write $u = 5\cdot k +1$. for some constant $k$ – The Integrator Jun 13 '18 at 17:58
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Then there will be two integers K and l – user568963 Jun 13 '18 at 18:01
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Then there will be two integers K and m. How would I find their value? – user568963 Jun 13 '18 at 18:01
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plug them back into the equation and for different values you'll get different numbers – The Integrator Jun 13 '18 at 18:03
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Given the conditions on $u$ and $v$, we can find $m$ and $n$ such that \begin{align*} u &= 5m + 1 \\ v &= 9n + 1. \end{align*} Substituting in, we get $$5m + 9n = 198$$ From here, we get a few answers. For example, $n = 2$ and $m = 36$, yielding $u = 181$ and $v = 19$. See if you can find some others.
Theo Bendit
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It gets even easier if you make this $10m = 396-18n$, so you need a multiple of $18$ which ends in $6$, with $n=2$ jumping out as the simplest example. You do not have to look at cases larger than $n=\frac{396}{18}$ – Henry Jun 13 '18 at 18:05
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One must be of the form $5r+1$ and the other of the form $9s+1$. Thus you wish to solve $5r+1+9s+1=200$ or $5r+9s=198$. The solution is not unique unless you add more constraints. For example the numbers $181$ and $19$ add to $200$ and have the property that you can divide $181$ by $5$ and $19$ by $9$ and get remainders of $1.$ The numbers $126$ and $64$ also have this property.
user1390
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