I cant figure out the way to solve this question,
Let n be a positive integer and {d1,d2,...,dr} be the whole positive divisor of n. Show that enter image description here then holds.
For example, when n = 12, φ(1) + φ(2) + φ(3) + φ(4) + φ(6) + φ(12) = 1 + 1 + 2 + 2 + 2 + 4 = 12 It certainly holds.
This statement is equivalent to proving that: $$\sum_{d\mid n}\varphi(d)=n$$ While there are many well-known proofs of this statement, the way I like to prove this is to first consider the fractions: $$\frac1{n},\frac{2}{n},\frac{3}{n},\cdots,\frac{n}{n}$$ Given that the numerators of the fractions run from $1$ to $n$, there are $n$ of these fractions. By expressing each fraction in its lowest term such that all fractions are in the form $\frac{p}{q}$ where $\gcd(p,q)=1$, the denominators of these new simplified fractions are all divisors of $n$.
The number of fractions which still have a denominator equal to $n$ is the number of fractions whose numerator was originally relatively prime to $n$, also denoted by $\varphi(n)$. Similarly, for any given divisor of $n$, which we will denote $d$, there are $\varphi(d)$ number of fractions with the denominator equal to $d$. The sum is therefore equal to the number of fractions, we have stated at the start as $n$. This proves the desired result.