There a infinity numbers of $n$ such that $ \phi (n) \equiv 0 \pmod{100} $

Question

I have no clue what this part: $ \phi (n) \equiv 0 \pmod {100} $ means. $0 \pmod {100}$ means I have an equivalence class $[0]$ in $\mathbb{Z}$. This also means I have $100, 200, 300 ,\cdots$ as the value of $ \phi (n)$.

Answer 1

All natural numbers n have a decomposition into prime factors. (Fundamental theorem of Arithmetic)

$n = p_1^rp_2^s\cdots p_{i}^z$

$\phi(n) = n (\frac {p_1 -1}{p_1})(\frac {p_2 -1}{p_2})\cdots(\frac {p_i -1}{p_i})$

If $100| \phi(n)$

Then $5^3$ must be a factor of $n$

In fact $\phi(5^3) = 4\cdot 5^2 = 100$

There exists at least one $n$ such that $100|\phi(n)$

For higher powers of $5,$ e.g. $\phi(5^4) = 500$

And for factor the factor 2, $\phi(2\cdot 5^3) = 2\phi(5^3)\cdot \frac 12 = 100$

$\forall j\ge 0, k\ge 3, \phi(2^j5^k) \equiv 0\pmod {100}$

Answer 2

You may take this as a Hint:

Consider this sequence, does it have this property?

$1000,10000,100000,1000000,\cdots$. Or equivalently, $10^k, k>2, k\in \mathbb{Z}.$ In fact, values of $\phi(10^k)$ are multiples of $400$,

can you show Why?

Answer 3

Since $\phi(ab)=\phi(a)\phi(b)$ for coprime $a,b$, it is sufficient to find some $a$ such that $\phi(a)=100$ and then all numbers $n=ab$ give $\phi(ab)\equiv 0\mod{100}$. Try $a=5^3$.