algebra

For discussing Olympiad Level Algebra (and Inequality) problems
Ridwan Abrar
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algebra

Unread post by Ridwan Abrar » Tue Aug 06, 2013 7:40 pm

Determine the number of positive integer solutions to the equation x+y+z+t=7

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Fahim Shahriar
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Re: algebra

Unread post by Fahim Shahriar » Sat Aug 10, 2013 12:41 am

When $x=1$ and $y=1$ , $z+t=5$ has $4$ solutions. Keeping $x=1$ fixed, $+1$ the value of $y$ gradually. Then $z+t$ will have $3,2$ and $1$ solutions for $y=2,3,4$ respectively.
So keeping $x=1$ fixed, we get $4+3+2+1=10$ solutions.

Keeping $x=2$ fixed, we get $(10-4)=6$ solutions. [4 solutions out. Because then z+t=5 is not possible].

Similarly, keeping $x=3,x=4$ fixed, we get $(6-3)=3$ and $(3-2)=1$ solutions respectively.

Total $= 10+6+3+1 = \boxed {20}$ $\text{solutions}$.
Last edited by *Mahi* on Fri Jan 24, 2014 11:09 am, edited 1 time in total.
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Name: Fahim Shahriar Shakkhor
Notre Dame College

Ridwan Abrar
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Joined:Mon Aug 05, 2013 8:01 pm

Re: algebra

Unread post by Ridwan Abrar » Thu Jan 23, 2014 11:48 pm

how do you understand that $z+t=5$ has 4 solutions?

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Phlembac Adib Hasan
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Re: algebra

Unread post by Phlembac Adib Hasan » Fri Jan 24, 2014 10:27 am

Ridwan Abrar wrote:how do you understand that z+t=5 has 4 solutions?
$1+4,2+3,3+2,4+1$. In fact $z+t=n$ has $n-1$ solutions in $\mathbb N$.
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