Happy factorizing

201400 may be factorized as $2^3 \times 5^2 \times 19 \times 53$ . Since we seek the most number, we treat this as simple $2 \times 2 \times 2 \times 5 \times 5 \times 19 \times 53$ , which has $\boxed{7}$ factors and thus $N$ has at most 7 distinct prime divisors.

Let N = p1^a1 p2^a2 .....*pn^an (p1,p2...,pn are distinct prime numbers and a1,a2... are natural numbers may or may not be equal.) Then 201400 = (a1+1)(a2+1).....(an+1)= 2×19×53×2×2×5×5 As on L.H.S. there are 7 factors; 7 is the answer. Note : a1,...,an are just powers not the primes so we can distribute those powers to atmost 7 distinct primes

In this question you need to find the # of distinct prime factors. Lets work backwards. The number of factors is found by (a+1)(b+1)(c+1)....where $2^{a}$ x $3^{b}$ x $5^{c}$ ... So all we need to do is factorize 201400 and find the number of prime factors.

When you factorize 201400, you get $2^{3}$ x $5^{2}$ x $19$ x $53$ . That means there are 7 (not distinct) prime factors of 201400 so there are 7 distinct prime factors of that number.

Note that $201400 = 2^{3} \cdot 5^{2} \cdot 19 \cdot 53$ . Since the number of divisors of an integer is the product of one more than each of the exponents in the prime factorization of the integer, there will be at most $\boxed{7}$ distinct prime divisors of $N$ .

Assume that the prime factorisation of $N$ is $p_1^{b_1}p_2^{b_2}p_3^{b_3}p_4^{b_4}...p_k^{b_k}$ , where $p_1,p_2,...,p_k$ are unique primes and $b_1,b_2,...,b_k$ are positive integers. Then the number of factors of $N$ can be written as:

$(b_1+1)(b_2+1)(b_3+1)...(b_k+1)=201400=19^1 \times 53^1 \times 2^3 \times 5^2$

Adding up the exponents of each prime factor of $201400$ gives us the maximum number of distinct primes that divide $N$ which is: $1+1+3+2=\boxed{7}$ .

If a number is divisible by a prime number, it means the prime number is in the prime decomposition of that number. Let the biggest number of primes that divide $N$ be $D$ .

$\Rightarrow N = x_{1}^{a_{1}} \times x_{2}^{a_{2}} \ldots \times x_{D}^{a_{D}}$ .

Then, remembering the tau formula for number of divisors: $t(N) = ( a_{1} +1 ) \times ( a_{2} +1 ) \ldots ( a_{D} +1 ) = 201400 = 2 \times 2 \times 2 \times 5 \times 5 \times 19 \times 53$

Now, we see that we want $D$ to be the biggest possible value. Therefore, we want the most amount of parenthesis possible.

Then, looking at the prime decomposition of $201400$ , we clearly see that the most amount of parenthesis possible is seven parenthesis (remember $a_{i} > 1$ for $i = 1 , 2 , \ldots , D$ ).

Therefore, the biggest possible value for $D$ is $7$ , so the most number of distinct primes which are divisors of $N$ are $\boxed{7}$ .

I'll try to express this:

If a number is factorized in the form $p^{a}_1 \cdot p^{b}_2 \cdot \ldots \cdot p^{k}_n$ , for some $k$ and $n$ (integers), then it's quantity of divisors is equal to $(a + 1)(b + 1)\ldots(k + 1)$ . Note that $201400 = 2014\cdot100 = 2\cdot2\cdot2\cdot5\cdot5\cdot19\cdot53$ . As it is given that $201400$ is the quantity of divisors, and maximizing, we say that the prime factorization of the number is $p^{1}_1 \cdot p^{1}_2 \cdot p^{1}_3 \cdot p^{4}_4 \cdot p^{4}_5 \cdot p^{18}_6 \cdot p^{52}_7$ . We've maximized, and we got $7$ different primes. Thus, our final answer is $\boxed {7}$ .