minimum value of Natural no. n

\hspace{-16}$Minimum value of $\bf{n\in\mathbb{N},}$ whic has ......\\\\ $\bf{(i)\;\; 16-}$ divisers.\\\\ $\bf{(ii)\;\; 19-}$ divisers.\\\\ $\bf{(iii)\;\; 24-}$ divisers.\\\\ $\bf{(iv)\;\; 25-}$ divisers.\\\\ $\bf{(v)\;\; 26-}$ divisers.\\\\

5 Answers

2305
Shaswata Roy ·

Is (i) 210.

1057
Ketan Chandak ·

1)210
2)218
3)25.33
4)24.34
5)212.3

Actually i am not sure about any of my answers... :P

2305
Shaswata Roy ·

1)210
2)218
3)25*3*5
4)24*34
5)212*3

If a number n can be factorized as p1n1p2n2......pknk(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.

1708
man111 singh ·

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

1708
man111 singh ·

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

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