which is greater ::
300^200 or 200 ^300 ??
which is greater ::
300^200 or 200 ^300 ??
200^300
HOW TO SOLVE DIS ONE ?? :(
Q3>Ravi who lives in the countryside ,caught a train for home earlier than usual , yesterday.Hs wife normally drives to the station to meet him,but yesterday he set out on foot from the station to meet his wife on the way.He reached home 12 mins earlier than he wud have done had he waited at the station for his wife .The car travels at a uniform speed , which is 5 times Ravi's speed on foot. Ravi reached home at exactly 6 o'clock.At what time wud he have reached home if his wife forewarne of his plan had met at the station?
(a)5:48 (b)5:24 (c)5:00 (d)5:36
ya my ans. also came out to be 200^300 but in arun sharma the answer is given as 300^200 is greater..so then the ans. in the book is wrong....???
the remainder when 10^10 + 10^100+10^1000+..........+10^(10^10) is divided by 7 is........???
asima Saysya my ans. also came out to be 200^300 but in arun sharma the answer is given as 300^200 is greater..so then the ans. in the book is wrong....???
Well... lemme check the explanation in the book... I'll confirm my answer then... Don't have the book just now...
deep_agrawal SaysWell... lemme check the explanation in the book... I'll confirm my answer then... Don't have the book just now...
I checked... And the answer for the question you have quoted is 200^300... Atleast in my book... Am i missing something here?
which is greater ::
300^200 or 200 ^300 ??
300^200=(3^200)*(2^400)*(5^400)
and 200 ^300=(2^900)*(5^600)
(300^200)/(200 ^300)=(3^200)/{(2^500)*(5^200)} which is surely
hence 200 ^300 > 300^200
What u say?
ohh:wow: i dont know then why in my book wrong ans. is given..thnk u
any one could post how to solve the next ques. that i had posted
asima Saysthe remainder when 10^10 + 10^100+10^1000+..........+10^(10^10) is divided by 7 is........???
according to Fermat's little theorem, 10^6%7 =1
so acc to that,
10^10%7=10^4%7 = 4 similarly,
10^100%7=10^4%7=4
10^1000%7= 10^4%7 =4
.
.
.
10^(10^10)%7= 10^4%7= 4
so the remainder is 4+4+.. 10 times= 40%7 = 5...
Hope this is correct. Pl revert back if this is wrong...
Reg,
Eshwar.
ya ans. is 5...
i understood how you have solved applying remainder theorem...but what is fermat's little theorem??
ya ans. is 5...
i understood how you have solved applying remainder theorem...but what is fermat's little theorem??
@asima, it wasn't remainder theorem that I used. But instead the Euler's theorem.. Pls revert back if there are some Qns...
I am pasting the extract from the post where I read about these,
FERMATS LITTLE THEOREM:
Using the properties of infinite arithmetic progressions, Fermat proved the theorem that for a prime number p that is co prime with another number a; when a to the power
(p-1) is divided by p, remainder is 1. Or a^(p-1) % = 1. Dont forget that this theorem is defined only for prime divisors. E.g. 2^4%5 =1; 3^10%11 = 1. Or in more general form
a^{n(p-1)}%p = 1. So if the question is 2^1000%11, remainder will be 1 because power of 2 is a multiple of (11-1).
What if divisor is not a prime? Then we have eulars generalization of Fermats rule.
That generalization says that a^e(n)%n = 1. Where e(n) is eulars number for divisor n. Also a is co prime with the divisor p. Eulars number e(n) for divisor n is defined as number of natural numbers less than n and co prime with it. How to find e(n)? Suppose I want to find eulars number of 1001. Prime factors of 1001 are 7,11,13.
e(1001) = 1001(1-1/7)(1-1/11)(1-1/13). In general for a number n having prime factors p1, p2, p3. e(n) = n (1-1/p1)(1-1/p2)(1-1/p3)
This knowledge about eulars theorem is sufficient.
If u have a strong hold of these 2 concepts then u can very well solve almost all remainder problems of the type a^n% b...
But there are some problems that cannot be solved, when the Euler's number is very large.. Just like 7^64%1000
here the euler number for 1000 is 100...
1000= 2^3*5^3
Euler of 2^3 =4 and euler of 5^3 = 100
so 7^4%8=1 and 7^100%125=1
taking LCM, we get 7^100%1000 = 1
(It has been mistakenly pointed out as 400 in some previous post)
so we have 7^100%1000 =1,
But we need to find 7^64%1000
There is a Euler's reverse or negative method(dont remember the name
exactly)... How to go about it was explained by Junnonmba in post (45-50), pl refer to those)For more on the Euler and Fermat, chk out http://www.pagalguy.com/297133-post73.html (mind blowing concepts@quant)
Reg,
Eshwar
can someone please solve chapter 9,lod 3,ques 9 on tsd
okk i got the ans.60 but my method will be bit long though its correct;
speed of 1st train=40;2nd =50;3rd lets assume=x;
let the 3rd train take t mins. to overtake the 1st train;then it will take(t+90)mins. to overtake the 2nd train;
thus we get two eqns.
1. 40*t=x*(t-30)
2. 50*(t+90)= x*(t-30+90)
solving the two eqns. we get t=90;
then putting the value of t=90 in either eqn.1 or eqn.2 we get x=60km/h(Ans.)
jha ji Sayscan someone please solve chapter 9,lod 3,ques 9 on tsd
Kind request to jha ji and asima... can u post the problem? ppl not having the book(like me) can also benefit...
Reg,
Esh
I have a question on permutation and combination..
How many ways can the letters of the word 'computer' be arranged so that vowels occupy the even positions?
It will be great if someone tells the way also how to solve it,rather than giving only the answer
regards,
Ravi Agnihotri
2 trains start simultaneously from the same point in the same direction.1st train speed is 40km/h & 2nd train's speed is 25%more than the 1st.
a third train starts from the same point & in the same dir. after 30 mins. if it overtakes the 2nd train 90 mins. later than it overtok the 1st train then what is the speed of the 3rd train?
Soln.
Speed of first train=40km/h
faster train=50km/hr
let the speed of third train is x
distance travelled in 30minutes by two trains will be 20km & 25 km
relative speed of third train wrt to 1st=x-40
for 2nd=x-50
25/(x-50) - 20/x-40=3/2
on solving x =60km/hr
2 trains start simultaneously from the same point in the same direction.1st train speed is 40km/h & 2nd train's speed is 25%more than the 1st.
a ..........
.....
TSD LOD III q-39 is also on the same lines
I have a question on permutation and combination..
How many ways can the letters of the word 'computer' be arranged so that vowels occupy the even positions?
It will be great if someone tells the way also how to solve it,rather than giving only the answer
regards,
Ravi Agnihotri
Soln.
No. of vowels =3
Non vowels =5
Even position= 4(2,4,6,
No. of ways of arranging 3 vowels in 4 places= 4P3=4!=24
after arranging no. of places left for non vowels =5
arraging of non vowels can be done in =5P5 ways= 5!=120
total no. of ways= 24*120=2880