2305Is (i) 210.
man111 singh actually Shaswata I dont have answerUpvote·0· Reply ·2013-02-24 03:47:20
\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.\\\\
-
UP 0 DOWN 0 0 5
5 Answers
2305
10571)210
2)218
3)25.33
4)24.34
5)212.3
Actually i am not sure about any of my answers... :P
- man111 singh Thanks Ketan may be some answers vary
23051)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.
- man111 singh Thanks Shaswata for Nice effort.
1708Should 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
- Shaswata Roy O yeah! Thanks for pointing that out.But I guess you understand the logic based on which I had answered the questions.Are they correct?
- Shaswata Roy O yeah! Thanks for pointing that out.But I guess you understand the logic based on which I had answered the questions.Are they correct?
1708Minimum 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
- man111 singh Yes Shaswata I have got the Concept.Thanks