Krdytkyu

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)$.

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

$n = p_1^rp_2^scdots p_{i}^z$

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

If $100| phi(n)$

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

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

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,
$phi(2cdot 5^3) = 2phi(5^3)cdot frac 12 = 100$

$forall jge 0, kge 3, phi(2^j5^k) equiv 0pmod {100}$

You may take this as a Hint:

Consider this sequence, does it have this property?

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

can you show Why?

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 $phi(ab)equiv 0mod{100}$. Try $a=5^3$.