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saifoo.khan
Questions coming..
awe some introduction waiting for it!
\(\tau(5)=2, \tau(7) = 2, \tau(12)=6,\tau(35)=4, \tau(6)=4, \tau(4)=3 \) So, for Q7. \(I, II\) are correct. For Q8, \(\tau (\tau (\tau(12))) = \tau(\tau(6)) = \tau(4) = 3\)
@FoolAroundMath , can you please explain what you wrote?
its 3 at question 8....
saif bhai i told you is type k questions nahi aye gai
first calculate the divisors for each of the no shown.... Divisors of 5: 1, 5 Divisors of 7: 1, 7 Divisors of 35: 1, 5, 7, 35 Divisors of 12: 1, 2, 3, 4, 6, 12 Divisors of 6: 1, 2, 3, 6
@wasiqss ; hehe. This is SAT 1.
and then count the no of divisors for each one then the ans by @FoolAroundMath follows
@wasiqss ; Most probably.
for \[\large n=\prod_{k=1}^{m} p_{k}^{\alpha_{k}} \\ \tau(n)=(1+\alpha_{1})(1+\alpha_{2})...(1+\alpha_{m})\] for example \[12=2^2*3^1 \\ \tau(12)=(2+1)(1+1)=6\]
@mukushla ; Sorry i don't get it . :D
@Raja99 , okay i get till the divisor thing.. what's next?
is it a way to fing the divisors @mukushla
then the value of the tau(5)=no of divisors of 5(Pl count it) and subsitute to chk the conditions.
for tau(5)=2 and tau(7)=2 and tau(35)=4 like that
and chk 2*2=4 then option is true...
Ah, nice., so; 1. 2 = 2 2. 2*2 = 4 3. 2+2 =/= 6 Perfect.
I was typing the whole thing out when OS crashed again. I'll leave it to the others to do that then ;)
@FoolAroundMath , Oh. okay. No Problem. (:
@Raja99 : Moving on to Question 8?
Divisors of 12: 1, 2, 3, 4, 6, 12 Divisors of 6: 1, 2, 3, 6 Divisors of 4: 1, 2, 4 ok then tau(12)=6 then tau(6)=4 then tau(4)=3 there fore ans is 3 option c
Perfect man! Thanks a ton. @Raja99
@saifoo.khan what math is this?
@lgbasallote : SAT.
correct me if im wrong but isnt SAT..uhmm a test?
Scholastic Aptitude "Test" @lgbasallote