There are three cities $a,b,c$. Each of these cities is connected
with the other two cities by at least one direct road. If a traveler
wants to go from one city to another city, she can do so either
by traversing a road connecting the two cities directly or by
traversing two roads, the first connecting the origin to the third
city and the second connecting the third city to destination. In all,
there are $33$ routes from $a$ to $b$, including through $c$. Similarly,
there are $23$ routes from $b$ to $c$, including $a$. How many roads
are there from $a$ to $c$ directly?
$\color{green}{a.)\ 6} \\
b.)\ 3 \\
c.)\ 5 \\
d.)\ 10 $
From the question I get
( $a$ to $b$ +$\ a$ to $c$ to $b$ ) $\ n(ab)+n(bc)+n(ac)=33$
($b$ to $c$ +$\ b$ to $a$ to $c$) $\ n(ab)+n(bc)+n(ac)=23$
I don't know how to proceed
I look for a short and simple way.
I have studied maths up to $12$ grade
Best Answer
Let $x$ be the number of roads from $a$ to $b$, $y$ be the number from $a$ to $c$, and $z$ be the number from $b$ to $c$. What you're told, when expressed mathematically, is that
$$x+yz=33$$ and $$z+xy=23$$
and you're asked to find $y$. If you subtract the second equation from the first, you get
$$(z-x)(y-1)=10$$
which implies $y=2$, $3$, $6$, or $11$. On the other hand, adding the two equations gives
$$(x+z)(y+1)=56$$
and this limits $y$ to be either $3$ or $6$, since $56$ is not divisible by $2+1=3$ or $11+1=12$. However, $y=3$ implies $z-x=5$ and $x+z=14$, which cannot be simultaneously satisfied by integers, since $x$ and $z$ need to have opposite parity to satisfy the first and equal parity to satisfy the second. So that leaves $y=6$ as the only possibility, with $z-x=2$ and $x+z=8$ satisfied by $x=3$ and $z=5$.