The number of positive integers n in the set | vedclass

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(B) fraction $\frac{1}{n}$ has a terminating decimal expansion if and only if the prime factorization of the denominator $n$ is of the form $2^a 	ime

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) fraction $\frac{1}{n}$ has a terminating decimal expansion if and only if the prime factorization of the denominator $n$ is of the form $2^a 	imes 5^b$,where $a, b \ge 0$ and $a+b > 0$.We need to find the number of integers $n \in \{2, 3, \ldots, 200\}$ of the form $2^a 	imes 5^b$.Possible values for $n$ are:- $2^a$: $2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32, 2^6=64, 2^7=128$- $5^b$: $5^1=5, 5^2=25, 5^3=125$- $2^a 	imes 5^b$: - $b=1$: $2^1 	imes 5^1=10, 2^2 	imes 5^1=20, 2^3 	imes 5^1=40, 2^4 	imes 5^1=80, 2^5 	imes 5^1=160$- $b=2$: $2^1 	imes 5^2=50, 2^2 	imes 5^2=100, 2^3 	imes 5^2=200$Listing all unique values: $2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, 100, 125, 128, 160, 200$.Counting these,we get a total of $18$ values.

The number of positive integers $n$ in the set $\{2, 3, \ldots, 200\}$ such that $\frac{1}{n}$ has a terminating decimal expansion is

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