Rachel does jumps, each of which is $168 \mathrm{~cm}$ long. Joel does jumps, each of which is $120 \mathrm{~cm}$ long. Mark does jumps, each of which is $72 \mathrm{~cm}$ long.

(a) On Monday, Rachel completes 5 jumps and Joel completes $n$ jumps. Rachel and Joel jump the same total distance. Determine the value of $n$.

(b) On Tuesday, Joel completes $r$ jumps and Mark completes $t$ jumps. Joel and Mark jump the same total distance, and $r$ and $t$ are integers. If $11 \leq t \leq 19$, determine the values of $r$ and $t$.

(c) On Wednesday, Rachel completes $a$ jumps, Joel completes $b$ jumps, and Mark completes $c$ jumps. Each of $a, b$ and $c$ is a positive integer, and Rachel, Joel and Mark each jump the same total distance. Determine the minimum possible value of $c$ and explain why this value is indeed the minimum.
(a) Since each of Rachel's jumps is $168 \mathrm{~cm}$ long, then when Rachel completes 5 jumps, she jumps $5 \times 168 \mathrm{~cm}=840 \mathrm{~cm}$.

Since each of Joel's jumps is $120 \mathrm{~cm}$ long, then when Joel completes $n$ jumps, he jumps $120 \mathrm{~cm}$.

Since Rachel and Joel jump the same total distance, then $120 n=840$ and so $n=7$.

(b) Since each of Joel's jumps is $120 \mathrm{~cm}$ long, then when Joel completes $r$ jumps, he jumps $120 \mathrm{r} \mathrm{cm}$.

Since each of Mark's jumps is $72 \mathrm{~cm}$ long, then when Mark completes $t$ jumps, he jumps $72 t \mathrm{~cm}$.

Since Joel and Mark jump the same total distance, then $120 r=72 t$ and so dividing by 24 , $5 r=3 t$.

Since $5 r$ is a multiple of 5 , then $3 t$ must also be a multiple of 5 , which means that $t$ is a multiple of 5 .

Since $11 \leq t \leq 19$ and $t$ is a multiple of 5 , then $t=15$.

Since $t=15$, then $5 r=3 \cdot 15=45$ and so $r=9$.

Therefore, $r=9$ and $t=15$.

(c) When Rachel completes $a$ jumps, she jumps $168 a \mathrm{~cm}$.

When Joel completes $b$ jumps, he jumps $120 b \mathrm{~cm}$.

When Mark completes $c$ jumps, he jumps $72 c \mathrm{~cm}$.

Since Rachel, Joel and Mark all jump the same total distance, then $168 a=120 b=72 c$.

Dividing by 24 , we obtain $7 a=5 b=3 c$.

Since $7 a$ is divisible by 7 , then $3 c$ is divisible by 7 , which means that $c$ is divisible by 7 . Since $5 b$ is divisible by 5 , then $3 c$ is divisible by 5 , which means that $c$ is divisible by 5 . Since $c$ is divisible by 5 and by 7 and because 5 and 7 have no common divisor larger than 1 , then $c$ must be divisible by $5 \cdot 7$ which equals 35 .

Since $c$ is divisible by 35 and $c$ is a positive integer, then $c \geq 35$.

We note that if $c=35$, then $3 c=105$ and since $7 a=5 b=105$, we obtain $a=15$ and $b=21$. In other words, $c=35$ is possible.

Therefore, the minimum possible value of $c$ is $c=35$.