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Let $S$ be the set of distinct values of $b$ for which the linear system $x+y+z=1$, $x+ay+z=1$, $ax+by+z=0$ has no solution. Describe the set $S$.
Subtract equation 1 from equation 2: $(a-1)y = 0$.
Case $a\neq1$: $y=0$. Then from eq.1 and eq.3: $x+z=1$ and $ax+z=0$, giving $(a-1)x=-1$, so $x=\frac{-1}{a-1}$. This always has a solution for $a\neq1$.
Case $a=1$: System becomes $x+y+z=1$, $x+y+z=1$, $x+by+z=0$. The third equation forces $-1=by-... $ i.e. subtracting eq.1: $(b-1)y=-1$. For no solution we need $b=1$.
So $S=\{1\}$ — a single value. Hence $S$ is a singleton.
If $\dfrac{1}{2^{11} \cdot 5^{17}}$ is written as a terminating decimal, how many nonzero digits will it contain?
Rewrite the expression to have a denominator that is a power of 10:
$\dfrac{1}{2^{11} \cdot 5^{17}} = \dfrac{1}{2^{11} \cdot 5^{17}} \times \dfrac{2^6}{2^6} = \dfrac{2^6}{2^{17} \cdot 5^{17}} = \dfrac{64}{10^{17}}$
So the decimal is $\dfrac{64}{10^{17}} = 0.\underbrace{00\ldots0}_{15}64$
The nonzero digits are $6$ and $4$ — that is, two nonzero digits.
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