Whodunni's Dice

Grumpelina, the Great Whodunni's beautiful assistant, placed a blindfold over the eyes of the famous stage magician. A member of the audience then rolled three dice.

'Multiply the number on the first dice by 2 and add 5,' said Whodunni. 'Then multiply the result by 5 and add the number on the second dice. Finally, multiply the result by 10 and add the number on the third dice.'

As he spoke, Grumpelina chalked up the sums on a blackboard which was turned to face the audience so that Whodunni could not have seen it, even if the blindfold had been transparent.

'What do you get?' Whodunni asked.
'Seven hundred and sixty three,' said Grumpelina.

Whodunni made strange passes in the air. 'Then the dice were '

What? And how did he do it?

(Source: Professor Stewart's Hoard of Mathematical Treasures by Ian Stewart)

Let the faces shown on the cast dice be represented as $d_1$, $d_2$ and $d_3$, respectively, and let the results called for by Whodunni be represented as $r_1$, $r_2$ and $r_3$, respectively. Then:

\begin{align*}
r_1 &= 2 \times d_1 + 5 \\
r_2 &= 5 \times r_1 + d_2 \\
r_3 &= 10 \times r_2 + d_3 = 763
\end{align*}

An assumption we seem warranted in making here is that the faces on the dice all have integer values. I will also assume, provisionally, that they are of the standard cubic variety (so-called d6s). If that is so, then $r_2$ can only be an integer if $d_3$ is three. If so, then $r_2$ is 76. Then $5 \times r_1 + d_2 = 76$. There are two possible ways for this equation to be satisfied, at first glance. $d_2$ could be 6. Then $r_1$ would have to be 14. This creates a problem. $2 \times d_1 + 5 = 14$, only possible with a non-integer solution for $d_1$. The other possibility is that $d_2$ is one. In this case, $r_1$ is 15. This is actually viable, because it means $d_1$ is five.

All told, the values on the dice were five, one and three, respectively.

In my opinion, Stewart's approach to solving this is even more interesting. Note that the calculation produces the numbers:

$$
2a + 5 \\
5(2a + 5)b \  (\text{or} \ 10a + b + 25) \\
10(10a + b + 25) + c \ (\text{or} \ 100a + 10b + c + 250)
$$

Subtracting 250 from their total, already known to be 763, results in 513. Determining the faces of the dice is then simply a matter of knowing how place value functions.