Question

If \(n\) is a positive integer, let \(S(n)\) denote the sum of the positive divisors of \(n\), including \(n\) itself, and let \(G(n)\) be the greatest divisor of \(n\). If \(H(n) = \dfrac{G(n)}{S(n)}\), then which of the following is the largest?

• A. \(H(2009)\)
• B. \(H(2010)\)
• C. \(H(2011)\)
• D. \(H(2012)\)

Hint:

The greatest divisor of any n is n itself, so H(n) = n/S(n) is largest when n has the fewest extra divisors, which happens when n is prime.

Answer 1

The Correct Option is C

Step 1: Rewrite H(n) using a reciprocal trick.
The greatest divisor of $n$ is $n$ itself, so $G(n) = n$ and $H(n) = n / S(n)$. Divide the numerator and denominator by $n$:
\[ H(n) = \frac{1}{S(n)/n} = \frac{1}{\sum_{d \mid n} \frac{1}{d}} \]
since if $d$ runs over every divisor of $n$, so does $n/d$, which means $S(n)/n$ is just the sum of $1/d$ over every divisor $d$ of $n$. So H(n) is largest exactly when this reciprocal sum is smallest.

Step 2: Figure out which kind of number makes that reciprocal sum small.
Every number has 1 and itself as divisors, contributing $1 + 1/n$ to the sum. Any extra divisor $d$ with $1 < d < n$ adds a further positive term $1/d$ on top of that. A number with no divisors besides 1 and itself, that is, a prime number, has the smallest possible reciprocal sum. Composite numbers always add extra positive terms and push the sum up.

Step 3: Check which of 2009, 2010, 2011, 2012 is prime.
2010 is even, so it is divisible by 2 right away, hence composite.
2009 factors as $7 \times 287 = 7 \times 7 \times 41$, so it is composite.
2012 is even too, in fact $2012 = 4 \times 503$, so it is composite.
2011 is odd, its digit sum is 4 so it is not divisible by 3, it does not end in 0 or 5, and testing every prime up to $\sqrt{2011} \approx 44.8$ (7, 11, 13, up to 43) leaves no exact division, so 2011 is prime.

Step 4: Conclude directly from the reciprocal-sum rule.
Since 2011 is prime, its reciprocal sum is just $1 + \frac{1}{2011} \approx 1.0005$, the smallest of the four choices by a wide margin. 2009, 2010 and 2012 are all composite and carry extra divisors like 2, 3, 7 or 41, each adding a noticeably larger extra term (for example $1/2 = 0.5$ for 2010, or $1/7 \approx 0.143$ for 2009), so their reciprocal sums sit well above 1.0005.

Final Answer:
The smallest reciprocal-divisor sum belongs to 2011, so H(2011) is the largest of the four.
\[ \boxed{\text{Option (C): } H(2011)} \]