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: Read what G(n) and H(n) actually mean.
The greatest divisor of any number \(n\) is \(n\) itself, since \(n\) always divides itself and no divisor of \(n\) can exceed \(n\). So \(G(n) = n\), and \(H(n) = \dfrac{n}{S(n)}\), where \(S(n)\) is the sum of all positive divisors of \(n\).

Step 2: Work out when H(n) is large.
Since \(H(n) = \dfrac{n}{S(n)}\), the fraction gets bigger as \(S(n)\) gets closer to \(n\). The smallest \(S(n)\) can be, for any \(n > 1\), is \(n + 1\), and that only happens when \(n\) has just two divisors, 1 and itself, which means \(n\) is prime. Any extra divisor adds more to \(S(n)\) without changing \(n\), so \(H(n)\) gets pulled down whenever \(n\) is composite.

Step 3: Factor each of the four numbers.
\(2009 = 7^2 \times 41\)
\(2010 = 2 \times 3 \times 5 \times 67\)
\(2011\) has no factor among any prime up to \(\sqrt{2011} \approx 44.8\) (checking 2, 3, 5, 7, 11, right up to 43, none divide it exactly), so \(2011\) is prime.
\(2012 = 2^2 \times 503\)

Step 4: Compute S(n) for each using the divisor-sum formula.
For \(n = p^a q^b\), \(S(n) = (1 + p + \dots + p^a)(1 + q + \dots + q^b)\).
\(S(2009) = (1 + 7 + 49)(1 + 41) = 57 \times 42 = 2394\)
\(S(2010) = (1+2)(1+3)(1+5)(1+67) = 3 \times 4 \times 6 \times 68 = 4896\)
\(S(2011) = 1 + 2011 = 2012\), since 2011 is prime and has only two divisors
\(S(2012) = (1 + 2 + 4)(1 + 503) = 7 \times 504 = 3528\)

Step 5: Compute each H(n) and compare.
\(H(2009) = \dfrac{2009}{2394} \approx 0.839\)
\(H(2010) = \dfrac{2010}{4896} \approx 0.410\)
\(H(2011) = \dfrac{2011}{2012} \approx 0.9995\)
\(H(2012) = \dfrac{2012}{3528} \approx 0.570\)
\(H(2011)\) is far bigger than the rest because 2011 being prime keeps \(S(n)\) as close to \(n\) as it can possibly get. The other three numbers each carry several small divisors (2, 3, 7, 41, and so on) that inflate \(S(n)\) well above \(n\), which drags their \(H(n)\) values down.

Final Answer:
\(H(2011)\) is the largest, so the answer is option (C). \[ \boxed{H(2011) = \dfrac{2011}{2012} \approx 0.9995} \]