Possible Duplicate:
Two algebra questions
How to approach the below question:
How many single-digit even natural number solutions are there for the equation $a+b+c+d = 24$ such that $a+b > c+d$?
Possible Duplicate:
Two algebra questions
How to approach the below question:
How many single-digit even natural number solutions are there for the equation $a+b+c+d = 24$ such that $a+b > c+d$?
The single-digit even natural numbers are $2$, $4$, $6$, $8$. The sum is $24$, quite big. So the number of possibilities is not large. Almost any careful listing will do the job. But here is a possible systematic approach.
The average of our numbers must be $6$. If they are all $6$, we violate $a+b\gt c+d$.
So there is at least one $8$. There can't be three $8$'s, that makes the sum too big. So the number of $8$'s is $1$ or $2$.
Now listing should be straightforward. Do (i) one $8$ and (ii) two $8$'s.
(i) There is only one $8$, and the average is $6$, so we must have one $8$, two $6$'s, and one $4$. Where can they be? Because $a+b\gt c+d$, one of $a$ or $b$ must be $8$, and the other $6$. And therefore one of $c$ or $d$ is $6$ and the other $4$. List all cases. You should get $4$.
(ii) All yours!
Here is one approach you can use:
The even single-digit (presumably in base 10) natural numbers are precisely 2, 4, 6, 8.
Thus, if $y$ and $z$ are such numbers, then $4\leq y+z\leq 16$, and of course $y+z$ is even.
Break the problem up as follows: find the number of solutions $w$, $x$ to $w+x=24$ such that $w$ and $x$ are even numbers such that $4\leq x
Here's a brute force approach:
L = (2,4,6,8) count = 0 for a in L: for b in L: for c in L: for d in L: if (a+b+c+d)==24 and (a+b)>(c+d): count = count + 1 print count
Which prints 11.