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 You are given the​ numbers, StartSet 36 plus n comma StartFraction n Over 3 EndFraction comma StartRoot n plus 16 EndRoot EndSet

. Find the smallest value of n so that all of the numbers in the set are natural numbers.

1 answer

We need to find the smallest value of \( n \) such that the expressions \( 36 + n \), \( \frac{n}{3} \), and \( \sqrt{n + 16} \) are all natural numbers.

  1. Analyze \( 36 + n \):

    • For \( 36 + n \) to be a natural number, \( n \) must be a natural number (which it will be).

  2. Analyze \( \frac{n}{3} \):

    • \( \frac{n}{3} \) is a natural number if \( n \) is a multiple of 3. Thus, let \( n = 3k \) for some natural number \( k \).

  3. Analyze \( \sqrt{n + 16} \):

    • For \( \sqrt{n + 16} \) to be a natural number, \( n + 16 \) must be a perfect square. Let \( n + 16 = m^2 \) for some integer \( m \). Then we have: \[ n = m^2 - 16 \]

  4. Setting \( n \) in terms of \( k \):

    • From our earlier deduction, since \( n = 3k \), we can substitute: \[ 3k = m^2 - 16 \quad \Rightarrow \quad m^2 = 3k + 16 \]

  5. Finding values for \( k \) and \( m \):

    • Rearranging gives us: \[ m^2 - 16 \equiv 0 \pmod{3} \]
    • Since \( 16 \equiv 1 \pmod{3} \), this simplifies to: \[ m^2 \equiv 1 \pmod{3} \]
    • The values of \( m \) which satisfy this condition are \( m \equiv 1 \pmod{3} \) or \( m \equiv 2 \pmod{3} \).

  6. Finding \( n \):

    • We will try some integer values for \( m \) (starting from 5, since \( m^2 - 16 \) should be positive):
      • If \( m = 5 \): \[ n = 5^2 - 16 = 25 - 16 = 9 \quad (9 \text{ is a multiple of } 3) \]
      • If \( m = 6 \): \[ n = 6^2 - 16 = 36 - 16 = 20 \quad (20 \text{ is NOT a multiple of } 3) \]
      • If \( m = 7 \): \[ n = 7^2 - 16 = 49 - 16 = 33 \quad (33 \text{ is a multiple of } 3) \]
      • If \( m = 8 \): \[ n = 8^2 - 16 = 64 - 16 = 48 \quad (48 \text{ is a multiple of } 3) \]

  7. Confirming the smallest \( n \):

    • We found \( n = 9 \) from \( m = 5 \).
    • Check if all conditions are satisfied for \( n = 9 \): \[ 36 + n = 36 + 9 = 45 \quad (\text{natural number}), \] \[ \frac{n}{3} = \frac{9}{3} = 3 \quad (\text{natural number}), \] \[ \sqrt{n + 16} = \sqrt{9 + 16} = \sqrt{25} = 5 \quad (\text{natural number}). \]

Thus, the smallest value of \( n \) such that all the expressions are natural numbers is:

\[ \boxed{9} \]