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**delicate_tears** · December 22nd 2007, 04:00 AM

from the Australian Intermediate Mathematics Olympiad:

**Soroban** · December 23rd 2007, 03:11 AM

Hello, delicate_tears!

Let $S,\,T\,,F$ = number of 6-, 10-, and 15-passenger taxis, respectively,
. . where: . $S,\,T,\,F\:\leq\:6$

We have: . $6S + 10T + 15F \:=\:120$

. . Hence: . $F \:=\:\frac{120-6S - 10T}{15} \quad\Rightarrow\quad F \:=\;8\; -\; \frac{2S}{5}\;-\; \frac{2T}{3}$ .[1]

Since $F$ is an integer, . $\begin{array}{cc}S\text{ is a multiple of 5:} & S = 5h \\ T\text{ is a multiple of 3:} & T = 3k \end{array}$

. . Then [1] becomes: . $F \;=\;8 - 2h - 2k \;=\;8 - 2(h + k)$

Since $S,\,T\:\leq\:6$ , then: . $h \:=\:0,1\text{ and }k \:=\:0,1,2$

It seems that there are: . $2 \times 3 \:=\:6$ possible choices.

However, if $h = k = 0$ , then $F = 8$ . . . not allowed.

Therefore, there are $\bf{\color{blue}5}$ ways.

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