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You are given the numbers {32+n,n/8,n+225}. Find the smallest value of n so that all of the numbers in the set are natural numbers.

Respuesta :

Ilyes39
n=0,8,16,24,...; because when n=8k [tex](k\in \mathbb{N})[/tex] then n/8 is natural number.
So the smallest value of n is 0.

As per definition of the natural numbers,

  • Set of all positive whole numbers are the natural numbers.

        Example → {1, 2, 3, 4.......∞}

        Smallest natural number → 1

        0 is not a natural number.

In the given set of numbers [tex]\{32+n, \frac{n}{8}, n + 225\}[/tex].

These numbers will be the natural numbers (without fraction).

Since, the smallest natural number is 1,

Therefore, the least value of [tex]\frac{n}{8}[/tex] will be 1.

[tex]\frac{n}{8}=1[/tex]

[tex]n=8[/tex]

Therefore, the given set of numbers will be the set of natural numbers [tex]\{40,1,233\}[/tex] if [tex]n=8[/tex].

Learn more about the natural numbers,

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