minimum value of Natural no. n

Is (i) 210.

1)210
2)2¹⁸
3)2⁵*3*5
4)2⁴*3⁴
5)2¹²*3

If a number n can be factorized as p₁ⁿ¹p₂ⁿ²......p_k^nk(where p₁,p₂...are all primes) then it has (n1+1)(n2+1).....(nk+1) factors.

Now if the number has say 6 factors then.

(n1+1)(n2+1)(n3+1)....(nk+1) = 6.

On factorizing we find that 6 = 2*3 = 6*1

therefore n1 and n2 are 1,2 or 5,0.

We find that 2¹*3²<2⁵

Hence the minimum value of n which has 6 divisors is 2¹*3².

Should it not be

The minimum value of n which has 6 divisors is = 12

bcz 6 = 6 * 1 = 2*3.

we can write it as = (5+1)*(0+1) = (0+1)*(5+1) = (1+1)*(2+1)=(2+1)*(1+1)

So for (5+1)*(0+1)= 2⁵*3⁰ = 32

So for (0+1)*(5+1) = 2⁰*3⁵ = 243

so for (1+1)*(2+1) = 2¹*3² = 18

So for (2+1)*(1+1) = 2²*3¹ = 12

So Min. Natural no. n which has 6 Divisers is = 12

Minimum value of Natural no. n which has 16 divisers is = 120(Which i have Got)

Solution:: 16 = 16*1 = 8*2 = 4*4 = 2*2*2*2 = 4*2*2

We can Write it as = (15+1)*(0+1) = (7+1)*(1+1) = (3+1)*(3+1) = (1+1)*(1+1)*(1+1)*(1+1)=(3+1)*(1+1)*(1+1)

So for first (15+1)*(0+1) = 2¹⁵

Similarly for (0+1)*(15+1) = 3¹⁵

So for (7+1)*(1+1) = 2⁷*3¹.

Similarly for (1+1)*(7+1) = 2¹*3⁷

So for (3+1)*(3+1) = 2³*3³

So for (1+1)*(1+1)*(1+1)*(1+1) = 2²*3²*5²*7²

So for (3+1)*(1+1)*(1+1) = 2³*3¹*5¹ = 120

So Minimum Natural no. n which has 16 Divisers is = 120