For a certain base $b$, the product $(12_b)(15_b)(16_b)$ is equal to $3146_b$. Let $s = 12_b + 15_b + 16_b$. What is $s$ in base $b$? I have worked out the above problem in the following way:
$(12_b)(15_b)(16_b)=(b+2)(b+5)(b+6)=3146_b=3b^3+b^2+4b+6$
$=>b^3+13b^2+52b+60=3b^3+b^2+4b+6$
$=>-2b^3+12b^2+48b+54=0$
$=>-b^3+6b^2+24b+27=0$
$=>6b^2+24b+27=b^3$
=>$2b^2+8b+9=b^3$
We need to find $s=3b+13$. I am stuck there. Any hint on how to proceed would be appreciated.
 A: Well done, until $6b^2+24b+27=b^3 \implies 2b^2+8b+9=b^3$. This step is wrong.
From $6b^2+24b+27=b^3$ we get that $b$ is a multiple of $3$. Since $6$ appears in $16_b$, we have $b>6$.
By inspection, $b=9$ is a root of $6b^2+24b+27=b^3$.
A: We have $$(b+2)(b+5)(b+6)=b^3+13b^2+52b+60=3b^3+b^2+4b+6$$
It follows the equation $$b^3-6b^2-24b-27=(b-9)(b^2+3b+3)=0$$
Hence $b=9$ and $$11+14+15=\color{red}{40}$$
A: Deciding that $b=9$ can be accomplished without resorting to cubic equations, and the answer to the posed question, "What is $s$ in base $b$?" is $44$ ($40$ given in previous answers is the base $10$ representation of $s$).
First, since $6$ is a digit in base $b$, $b>6$.
Next, since $12\times 15\times 16=2880_{10}$, the fact that the representation in base $b$ is $3146$ means that $b<10$. If that isn't apparent, I will explain it after resolving the main question.
Finally, looking only at the units digits, $2\times 5\times 6=60_{10}$, but the units digit in the base $b$ representation is $6$. That means $54_{10}$ units have to be carried out of the units place to a higher place (the $b^2$ place, or possibly even the $b^3$ place). That further means that $54$ is a multiple of $b$
The only conclusion consistent with these facts is $b=9$, and checking the math, it works. Reverting to base $10$ to multiply conveniently, we get $11\times 14\times 15=2310$, which when converted to base $9$ is represented as $3146$.
As an aside, I note that $2310=2\cdot 3\cdot 5\cdot 7\cdot 11$ which is the product of the first $5$ primes, or the primorial $p_5\#$. Thus, $3146_9=p_5\#$.
Now to the understanding that a "larger" appearing representation of a quantity results from being presented in a smaller base. The best way to appreciate this is to look at a fixed quantity, say the quantity we mean when we say "one hundred," in many different bases. In what follows, I will use $t$ to indicate a digit with the value of ten, and $f$ to indicate a digit with the value of fifteen. $$\text{One hundred}= 50_{20}=55_{19}=5t_{18}=5f_{17}=64_{16}=6t_{15}=72_{14}=79_{13}=84_{12}=91_{11}=100_{10}=121_9=144_8=202_7=244_6=400_5=1210_4=10201_3=1100100_2$$
The smaller the base, the "larger" the representation of a fixed quantity appears.
