Discussion
Let \(\varphi(n)\) denote Euler's function. Then :
\[ \sum_{d|24} \varphi(d) \]
is equal to :
Answer: A
The sum is 24.
Gauss's identity says \(\sum_{d|n}\varphi(d) = n\) for every \(n\). Checking with the divisors of 24:
\[ \begin{aligned} &\varphi(1) + \varphi(2) + \varphi(3) + \varphi(4) \\ &+ \varphi(6) + \varphi(8) + \varphi(12) + \varphi(24) \\ &= 1 + 1 + 2 + 2 + 2 + 4 + 4 + 8 = 24 \end{aligned} \]
The reason is that each of the numbers 1 to 24 has a unique gcd \(g\) with 24, and those with gcd \(g\) number \(\varphi(24/g)\). 25 and 23 are off by one, and 20 omits terms.
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