Week 3 Solutions

Answer for wk 3:

1) 3628799 (Online source)
2) $\displaystyle{\frac{59}{3}}$ (2008 AMC 10A)

Solution (1):
We try to derive a general formula for $n\cdot n!$ . We can see that

$n\cdot n! = (n+1)!-n!$ , for every $n \in \mathbb{N}$ .

Don't ask me how I get this -- you have to practice lots of problems to see the pattern. Thus, applying the formula, we get

$1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots +9\cdot 9! = (2!-1!)+(3!-2!)+\cdots +(10!-9!)=10!-1=3628799$

Solution (2):
At first glance, this problem seems hard to solve because it only gives us the perimeter and area. However, notice that it is a right triangle. This inspires us to set the legs of triangle as a,b, and its hypotenuse as $\sqrt{a^2+b^2}$ . Thus, we arrive at the simultaneous equations:

$a+b+\sqrt{a^2+b^2}=32 \textup{ ------(1)}$
$\displaystyle{\frac{1}{2}ab=20} $ \textup{ -----(2)}$

Simplifying the first equation gives us:

$\sqrt{a^2+b^2}=32-(a+b)$
$a^2+b^2=32^2-64(a+b)+(a+b)^2$
$a^2+b^2=32^2-64(a+b)+a^2+b^2+2ab$
$64(a+b)=1024+2ab$

From here we can see that it is impossible to find the value of a,b respectively. Hence, we change our approach -- we find a+b and subtract it from the perimeter of the triangle, which is given, to obtain the hypotenuse. Therefore,

$\displaystyle{a+b=\frac{2ab+1024}{64}} \textup{ -----(3)}$

From (2),

$2ab=80\textup{ -----(4)}$

Substituting (4) into (3),

$\displaystyle{a+b=\frac{1104}{64}}$
$\displaystyle{a+b=\frac{69}{4}}$

Therefore, the length of the hypotenuse is $\displaystyle{p-a-b=32-\frac{69}{4}=\frac{59}{4}}$ .