Question 34

The smallest set of ages for the grandson is $\{1$, $2$, $3$, $4$, $5$, $6\}$.

Suppose the man was $A$ years old when the grandson was born. Then:

• $A+1$ is a multiple of $1$
• $A+2$ is a multiple of $2$
• $A+3$ is a multiple of $3$
• $A+4$ is a multiple of $4$
• $A+5$ is a multiple of $5$
• $A+6$ is a multiple of $6$.

Therefore $A$ is a multiple of $1$, $2$, $3$, $4$, $5$, and $6$.

Now the lowest common multiple of $1$, $2$, $3$, $4$, $5$, and $6$ is $60$.

$\therefore\enspace A$ must be a multiple of $60$.

$\therefore\enspace A = 60$

$\therefore\enspace$on the sixth of these birthdays, the man is $66$ and his grandson is $6$.

Note:

By Pythagoras in triangle ABC,

$\begin{aligned} (R-r)^{2} &= r^{2}+\left(\dfrac{R}{2}\right)^{2} \\ \\ \therefore\enspace R^{2}-2Rr+r^{2} &= r^{2}+\dfrac{R^{2}}{4} \\ \\ \therefore\enspace R^{2}-2Rr &= \dfrac{R^{2}}{4} \\ \\ \therefore\enspace 4R^{2}-8Rr &= R^{2} \\ \\ \therefore\enspace 3R^{2}-8Rr &= 0 \\ \\ \therefore\enspace R(3R-8r) &= 0 \\ \\ \therefore\enspace 3R &= 8r \quad\{R=0\} \\ \\ \therefore\enspace \dfrac{r}{R} &= \dfrac{3}{8} \\ \\ \therefore\enspace r:R &= 3:8 \end{aligned}$