# Tourists & Buses et al.

• September 11th 2007, 05:45 AM
puellaevixlegitimae
Tourists & Buses et al.
1) A group of tourists is offered seats in a number of buses so that there were the same number of tourists in each bus. First the organizers tried 22 tourists in each bus, but 1 was left unseated. Then 1 bus left empty and the tourists used seats in the remaining buses. Find the number of buses and tourists if each bus has only 44 seats.

2) 2 cyclists started to ride at 8 am, one from A to B, and the other from B to A. Each rode at a constant speed along the same road and when each arrived at the terminal point, immediately turned back. They met for the first time at 11 am and each of them turned exactly once before they met for a second time. Find the time of their second meeting.

3) 1999 numbers are placed around the circumference of a circle. When any four successive numbers are added, the total is always 28. What are these 1999 numbers? Find all possibilities.
• September 11th 2007, 07:30 AM
ThePerfectHacker
Quote:

Originally Posted by puellaevixlegitimae
1) A group of tourists is offered seats in a number of buses so that there were the same number of tourists in each bus. First the organizers tried 22 tourists in each bus, but 1 was left unseated. Then 1 bus left empty and the tourists used seats in the remaining buses. Find the number of buses and tourists if each bus has only 44 seats.

Let $t$ be the number of tourist, let $b$ be the number of buses, let $s$ be the number of seats.

From reading the problem if we had just one more tourist, i.e. $t+1$ we would be able to evenly divide them among the 22 seats so $t+1 = 22b$. The other equation says that if we had one bus less then the tourists can be evenly divided with $s$ seats in each both, so, $t=(b-1)s$.

Thus, we have two equations,
$\left\{ \begin{array}{c}t+1=22b \\ t = (b-1)s \end{array} \right.$

Substitute,
$(b-1)s+1=22b$
$bs-s+1=22b$
$-s+1 = 22b-bs$
$-s+1 = b(22-s)$
$s-1 = b(s - 22)$
The last equation tells us that $s > 22$ because the LHS is positive while RHS would not be.
Thus,
$b = \frac{s-1}{s-22}$.
Note that $b$ is an integer so the right hand side must be an integer.
Since by the conditions of the problem $s\leq 44$.
And the contrainst that we found we need to find all $s$ with $22 such that $\frac{s-1}{s-22}$ is an integer.
We can throw away half the cases if we realize that the denominator and numerator have opposite parity. And so the denominator cannot be even. That happen only when $s$ is even. Thus, we need to check the cases $s = 23,25,27,29,31,33,35,37,39,41,43$ by hand which is fast.
But I notice (without doing that) a trivial answer of $s=23$ for the makes the denominator 1.
Hence $s=23,b=22,t=483$.

Note: There might be other solutions which I did not check for.
• September 11th 2007, 07:58 AM
Soroban
Hello, puellaevixlegitimae!

There is a clever approach to #2 . . .

Quote:

2) Two cyclists started to ride at 8 am, one from $A$ to $B$, and the other from $B$ to $A$.
Each rode at a constant speed along the same road
and when each arrived at the terminal point, they immediately turned back.
They met for the first time at 11 am and each of them turned exactly once
before they met for a second time. .Find the time of their second meeting.

It helps if you make a sketch of their travelling.

Let $d$ be the distance between $A$ and $B.$

They started at 8 am. .At 11 am they met at $P$.
. . It took them $3$ hours to cover a total of $d$ miles.

They continue on their way, turn at the end, and cycle towards each other again.
. . Then they meet at point $Q$.
Their total distance (since 8 am) is $3d.$ .
. . To cover $3d$ miles, it takes them $9$ hours.
Therefore, their second meeting was at $5\text{ p.m.}$