The number of integers q,1 leq q leq 2021,suc | vedclass

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(B) For $\frac{1}{q}$ to have a terminating decimal expansion,$q$ must be of the form $2^m 5^n$ where $m, n \in \mathbb{W}$.For $\sqrt{q}$ to be ratio

Vedclass · Accepted Answer

$11$

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(B) For $\frac{1}{q}$ to have a terminating decimal expansion,$q$ must be of the form $2^m 5^n$ where $m, n \in \mathbb{W}$.For $\sqrt{q}$ to be rational,$q$ must be a perfect square.If $q = 2^m 5^n$ is a perfect square,then both $m$ and $n$ must be even.Let $m = 2a$ and $n = 2b$ where $a, b \in \mathbb{W}$.Then $q = 2^{2a} 5^{2b} = (2^a 5^b)^2$.We are given $1 \leq q \leq 2021$,so $1 \leq (2^a 5^b)^2 \leq 2021$,which implies $1 \leq 2^a 5^b \leq \sqrt{2021} \approx 44.95$.We list the possible values for $2^a 5^b \leq 44$:If $b=0$: $2^a \leq 44 \Rightarrow a \in \{0, 1, 2, 3, 4, 5\}$ (Values: $1, 2, 4, 8, 16, 32$)If $b=1$: $2^a \cdot 5 \leq 44$ $\Rightarrow 2^a \leq 8.8$ $\Rightarrow a \in \{0, 1, 2, 3\}$ (Values: $5, 10, 20, 40$)If $b=2$: $2^a \cdot 25 \leq 44$ $\Rightarrow 2^a \leq 1.76$ $\Rightarrow a = 0$ (Value: $25$)Total values for $2^a 5^b$ are $6 + 4 + 1 = 11$.Thus,there are $11$ such integers $q$.

The number of integers $q$,$1 \leq q \leq 2021$,such that $\sqrt{q}$ is rational and $\frac{1}{q}$ has a terminating decimal expansion,is

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