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n is an integer greater than 0.
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22 Jun 2025, 13:46
22 Jun 2025, 13:46
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\(\begin{aligned}
&n \text { is an integer greater than } 0 .\\
&\begin{array}{cc}
\text { Quantity A } & \text { Quantity B } \\
\text { Number of prime } & \text { Number of prime } \\
\text { factors of } 2 n(2 n+2) & \text { factors of } n(n+1)
\end{array}
\end{aligned}\)
&n \text { is an integer greater than } 0 .\\
&\begin{array}{cc}
\text { Quantity A } & \text { Quantity B } \\
\text { Number of prime } & \text { Number of prime } \\
\text { factors of } 2 n(2 n+2) & \text { factors of } n(n+1)
\end{array}
\end{aligned}\)
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Re: n is an integer greater than 0.
[#permalink]
24 Jun 2025, 04:15
24 Jun 2025, 04:15
Expert Reply
1. Simplify Quantity A:
Quantity A's expression is $2 n(2 n+2)$.
We can simplify it:
$$
2 n(2 n+2)=2 n \cdot 2(n+1)=4 n(n+1)=2^2 \cdot n(n+1)
$$
2. Analyze Distinct Prime Factors:
- Quantity B asks for the number of distinct prime factors of $\(n(n+1)\)$.
- Quantity A asks for the number of distinct prime factors of $\(2^2 \cdot n(n+1)\)$.
3. The Key Insight:
The product of two consecutive integers, $\(n(n+1)\)$, is always even. This means that 2 is always a distinct prime factor of $\(n(n+1)\)$.
Since 2 is already a distinct prime factor of $\(n(n+1)\)$, multiplying $\(n(n+1)\)$ by $\(2^2\)$ (as in Quantity A's expression) does not introduce any new distinct prime factors. It only increases the exponent of the existing prime factor 2 in the overall prime factorization.
Therefore, the set of distinct prime factors for both quantities is identical.
Conclusion:
The number of distinct prime factors for Quantity A is the same as for Quantity B.
The final answer is The two quantities are equal.
Quantity A's expression is $2 n(2 n+2)$.
We can simplify it:
$$
2 n(2 n+2)=2 n \cdot 2(n+1)=4 n(n+1)=2^2 \cdot n(n+1)
$$
2. Analyze Distinct Prime Factors:
- Quantity B asks for the number of distinct prime factors of $\(n(n+1)\)$.
- Quantity A asks for the number of distinct prime factors of $\(2^2 \cdot n(n+1)\)$.
3. The Key Insight:
The product of two consecutive integers, $\(n(n+1)\)$, is always even. This means that 2 is always a distinct prime factor of $\(n(n+1)\)$.
Since 2 is already a distinct prime factor of $\(n(n+1)\)$, multiplying $\(n(n+1)\)$ by $\(2^2\)$ (as in Quantity A's expression) does not introduce any new distinct prime factors. It only increases the exponent of the existing prime factor 2 in the overall prime factorization.
Therefore, the set of distinct prime factors for both quantities is identical.
Conclusion:
The number of distinct prime factors for Quantity A is the same as for Quantity B.
The final answer is The two quantities are equal.
gmatclubot
Re: n is an integer greater than 0. [#permalink]
24 Jun 2025, 04:15
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