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Question #1 (Question un ) : What is the smallest positive integer k such that k/660 can be expressed as a terminating decimal?

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Sagot :

Answer:

k = 33

Step-by-step explanation:

Terminating decimal numbers:  Decimals that have a finite number of decimal places.

For a decimal to be terminating, the factors of the denominator must only contain 2 and/or 5.  As 2 and 5 are prime numbers, use prime factorization to rewrite the denominator.

Prime factorization of 660:

⇒ 660 = 2 × 2 × 3 × 5 × 11

⇒ 660 = 2² × 3 × 5 × 11

Therefore:

[tex]\implies \sf \dfrac{k}{660}=\dfrac{k}{2^2 \cdot 3 \cdot 5 \cdot 11}[/tex]

The fraction will only be a terminating decimal if both 3 and 11 in the denominator are canceled out.  To do this, their lowest common multiple must be the numerator:

⇒ LCM of 3 and 11 = 3 × 11 = 33

[tex]\implies \sf \dfrac{33}{660}[/tex]

[tex]\implies \sf k=33[/tex]

Therefore, the smallest positive integer k such that k/660 can be expressed as a terminating decimal is 33.

Answer:

33

Step-by-step explanation:

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