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Use the Fundamental Theorem of Arithmetic to justify parts (a) and (b).


AIf $5 \mid n$ and $3 \mid n$, then $15 \mid n$.
BIf $9 \mid n$ and $25 \mid n$, then $225 \mid n$.
CIs it always true that if $a \mid n$ and $b \mid n$, then $ab \mid n ^2$? Either prove the statement or give a counterexample.


AThe Fundamental Theorem of Arithmetic states that $n$ can be written as a [ ] in one and only one way. Since $5 \mid n$ and $3 \mid n$, and since 5 and 3 are both [ ], they must be included in the unique [ ]. Since [ ] = [ ], $15 \mid n$.


BSince $9 \mid n$ and $25 \mid n$, and since 9 and 25 are [ ], they must be included in the unique [ ]. Since [ ] = [ ], $225 \mid n$.


CSelect the correct choice below and, if necessary, fill in the answer box within your choice.
ANo. As a counterexample, let $a = 2$, $b = 3$, and $n = 15$. Then, $n^2 = [ ]$, which is not divisible by both $a$ and $b$.
BNo. As a counterexample, let $a = 2$, $b = 3$, and $n = 6$. Then, $n^2 = [ ]$, which is not [ ].

Verified Answer :

  • The Fundamental Theorem of Arithmetic guarantees that integers have unique prime factorizations, which we use to show that if \(n\) is divisible by coprime factors \(a\) and \(b\), it is divisible by \(ab\).
  • For part (a), \(5 \mid n\) and \(3 \mid n\) implies \(15 \mid n\) because 5 and 3 are distinct primes.
  • For part (b), \(9 \mid n\) and \(25 \mid n\) implies \(225 \mid n\) because \(3^2\) and \(5^2\) are coprime.
  • For part (c), if \(a \mid n\) and \(b \mid n\), then \(n = ak = bm\). Squaring \(n\) gives \(n^2 = (ak)(bm) = ab(km)\), which proves that \(ab \mid n^2\) is always true.

Explanation

  1. Part (a) Analysis and Justification

The Fundamental Theorem of Arithmetic states that every integer \(n > 1\) can be written as a product of prime numbers in a unique way, except for the order of factors.

Part (a): If \(5 \mid n\) and \(3 \mid n\), then \(15 \mid n\).
Since \(5 \mid n\) and \(3 \mid n\), the prime factorization of \(n\) must include at least one factor of 5 and at least one factor of 3. Because 5 and 3 are distinct primes, their product, \(3 \times 5 = 15\), must also be a factor of \(n\) in its unique prime factorization. Thus, \(15 \mid n\). The completed sentences are:
The Fundamental Theorem of Arithmetic states that \(n\) can be written as a \(\boxed{\text{product of prime numbers}}\) in one and only one way. Since \(5 \mid n\) and \(3 \mid n\), and since 5 and 3 are both \(\boxed{\text{prime numbers}}\), they must be included in the unique \(\boxed{\text{prime factorization of } n}\). Since \(\boxed{3 \times 5} = \boxed{15}\), \(15 \mid n\).

  1. Part (b) Analysis and Justification

Part (b): If \(9 \mid n\) and \(25 \mid n\), then \(225 \mid n\).
We are given that \(9 \mid n\) and \(25 \mid n\). Since \(9 = 3^2\) and \(25 = 5^2\), for \(n\) to be divisible by 9, its prime factorization must contain \(3^2\). For \(n\) to be divisible by 25, its prime factorization must contain \(5^2\). Because 3 and 5 are distinct primes, the factors \(3^2\) and \(5^2\) are coprime (i.e., their greatest common divisor is 1). Therefore, \(n\) must be divisible by their product, \(9 \times 25 = 225\).
The completed sentences are:
Since \(9 \mid n\) and \(25 \mid n\), and since 9 and 25 are \(\boxed{\text{coprime}}\), they must be included in the unique \(\boxed{\text{prime factorization of } n}\). Since \(\boxed{9 \times 25} = \boxed{225}\), \(225 \mid n\).

  1. Part (c) Analysis and Validity

Part (c): Is it always true that if \(a \mid n\) and \(b \mid n\), then \(ab \mid n^2\)?
Let \(a \mid n\) and \(b \mid n\). This means there exist integers \(k\) and \(m\) such that \(n = ak\) and \(n = bm\).
If we multiply these two equations, we get \(n^2 = (ak)(bm) = (ab)(km)\).
Since \(k\) and \(m\) are integers, their product \(km\) is also an integer. This directly shows that \(n^2\) is a multiple of \(ab\), so \(ab \mid n^2\) is always true.
Because the statement is always true, the provided options A and B (which claim the statement is false and provide counterexamples) are incorrect. The premise of these choices is fundamentally flawed. In all cases where \(a \mid n\) and \(b \mid n\), \(ab \mid n^2\) holds.

Examples

Understanding the properties of divisors is crucial in fields like cryptography. For example, if you are designing a security protocol that requires finding numbers divisible by multiple factors, the Fundamental Theorem of Arithmetic ensures that you can always determine divisibility by products of coprime factors. This is a foundational concept for algorithms that check if a large number has specific prime components, helping to verify data integrity in digital communications.

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