YES MATE.The basic thing here in this case is the remainder theorem.
you and i are doing just the same thing in the problem,but just in a different way.
pls repeat your process with question what will be remainder when 33^34^35 divided by 7?
pls repeat your process with question what will be remainder when 33^34^35 divided by 7?
How many 3 digit positive integer with digits x,y and z in hundred ten and unit place respectively exist such that x
How many 3 digit positive integer with digits x,y and z in hundred ten and unit place respectively exist such that x
The number is of the form xyz. Y can have any value from 2 to 9 (not 1 as y has to be greater than x and z and x cant be 0)
When y = 9, X can have any value from 1 to 8 i.e 9 values and Z from 0 to 8 i.e 9 values => 8*1*9 possible numbers
Similarly when y = 8, X => 1 to 7 i.e 7 values and z => 0 to 7 i.e 8 values => 7*1*8 possible numbers
So if we check, for every value of z we can form n(n+1) numbers where n goes from 1 to 8
Hence the answer = summation of n(n+1), where n goes from 1 to 8
= summation of (n^2+n), where n goes from 1 to 8
= , where n goes from 1 to 8
= n(n+1)(2n+1)/6 + n(n+1)/2, n goes from 1 to 8
= 8*9*17/6 + 8*9/2
= 240
dark_knight89 Sayscan you please explain the format little theorem
Fermat's little theorem has been explained here
http://www.pagalguy.com/forum/cat-and-related-discussion/41201-ahmedabad-study-group-cat-2010-a-9.html#post1623914
LOD2 TIME,SPEED AND DISTANCE QUESTION33 ravi,who lives in the countryside,caught a train for home earlier than usual yesterday.his wife normally drive to the station to meet him.but yesterday he set out from the station to meet his wife on the way.he reached home 12 minutes earlier than he would have done had he waited at the station for his wife.the car travels at the uniform speed,which is five times the ravis speed on the foot.ravi reached home at exactly 6"0 clock.at what time would he have reached home if his wife ,forewarned of his plan,had met him at the station?a)5:48 b)5:24 c)5:00 d)5:36 ans(d)
QUESTION34 hemant and ajay start the two length swimming race at the same moment but from the opposite end of the pool.they swim in the lane and with uniform speeds,but hemant is faster than ajay.they first passed at point 18.5m from the deep end and having comleted the one length.each one is allowed to take a rest on edgr for exactly 45 seconds.after setting off on the return length,the swimmers pass for the second time just 10.5 m fro the shallow end.how long is the pool?
(a)55.5m (b)45m (c)66m (d)49m ans(b)
KINDLY SENT THE DETAILED SOLUTION
34Sol: Length of Pool = l
Ratio of speeds = (l-18.5)/18.5
Next time meeting point distances = 2l-10.5 and l+10.5
Time taken is same.
So, (l-18.5)/18.5=(2l-10.5)/(l+10.5)
=>37l-194.25=l^2-8l-194.25
=>l^2-45l=0
=>l=45m!
33rd, I tried but couldn't arrive at a solution 😞
This one in from AP/GP(LOD3-32)
In a certain colony of cancerous cells,each cell reproduces by giving birth to 2 new cells every hour. If there is a single productive cell at the start and this process continues for 9 hrs how many cells will the colony have at the end of 9 hrs.Life of an individual cell is 20hrs and only 50% of cells are capable to produce the next generation?
This one in from AP/GP(LOD3-32)
In a certain colony of cancerous cells,each cell reproduces by giving birth to 2 new cells every hour. If there is a single productive cell at the start and this process continues for 9 hrs how many cells will the colony have at the end of 9 hrs.Life of an individual cell is 20hrs and only 50% of cells are capable to produce the next generation?
Not sure if this solution is flawless or not.. but my approach is this
As 9hrs is less than life of individual cell, we need not worry about death of any cells. As only 50% cells capable to produce new generation => 1st cell {prodcutive} gives rise to 2 cells, out of these 2 cells one will be productive to give rise to a total of 1+2 cells.. Similarly, if u observe,
After 1 hour :1 -> 2
After 2 hours:2-> 1+2
After 3 hours:1+2-> 1+1+2
Similarly after 9 hours, we get 8*1+2 = 10 cells.
So answer is 10. Please confirm if it is right or wrong !
actually the answer is 2^9
Not sure if this solution is flawless or not.. but my approach is this
As 9hrs is less than life of individual cell, we need not worry about death of any cells. As only 50% cells capable to produce new generation => 1st cell {prodcutive} gives rise to 2 cells, out of these 2 cells one will be productive to give rise to a total of 1+2 cells.. Similarly, if u observe,
After 1 hour :1 -> 2
After 2 hours:2-> 1+2
After 3 hours:1+2-> 1+1+2
Similarly after 9 hours, we get 8*1+2 = 10 cells.
So answer is 10. Please confirm if it is right or wrong !
i also solved this question,i believe answer given in book is wrong
even i agree with you guys.. my answer has been 10 too
SHIKHA mehta Sayseven i agree with you guys.. my answer has been 10 too
well,ans is 2^9+1.
after 1 hour=>1+2 new cells=3
after2 hours=3+2 new cells=5=2^2+1(as only 50%r able to produce new cells.)
similarly after 3 hours=5+4=2^3+1
so after 9 hours=2^9+1..
hope u got it..
well,ans is 2^9+1.
after 1 hour=>1+2 new cells=3
after2 hours=3+2 new cells=5=2^2+1(as only 50%r able to produce new cells.)
similarly after 3 hours=5+4=2^3+1
so after 9 hours=2^9+1..
hope u got it..
1 cell gives rise to only 2 cells.. not 1+2 cells..
SHIKHA mehta Sayseven i agree with you guys.. my answer has been 10 too
there is lot of errors in Arun sharma. If ur confident with ur answer u can move ahead..
Number Systems, LOD2, Q 61
Define a no. K such that it is the sum of squares of the first M natural nos (k=1^2+2^2+3^2+....+M^2) where Moptions: 10, 11, 12
Any short cuts???:-(
Number Systems, LOD2, Q 61
Define a no. K such that it is the sum of squares of the first M natural nos (k=1^2+2^2+3^2+....+M^2) where Moptions: 10, 11, 12
Any short cuts???:-(
Is ans 12?
luckymurari Says1 cell gives rise to only 2 cells.. not 1+2 cells..
its total no of cells after 1 hour.
1 is parent and 2 newly formed..so total is 3.
b'coz life of an individual cell is 20 hours.
now hope u got it.
Number Systems, LOD2, Q 61
Define a no. K such that it is the sum of squares of the first M natural nos (k=1^2+2^2+3^2+....+M^2) where Moptions: 10, 11, 12
Any short cuts???:-(
correct me if I am wrong. I think the answer should be 13.
summation of n^2=n(n+1)(2n+1)/6. For this to be divisible by 4 ;
n(n+1)(2n+1) should have 2^3 and 3. Since only one the 3 factors can be even always, n should take values which are multiples of 7 and multiples of 8 only.
multiples of 8 multiples of 7 I feel it should be 13..
ans is 12...
abhisekjena Saysans is 12...
Ok..I get it now.. I was wrong in assuming that multiples of 7 also should be considered.
All the multiples of 8 and their previous number should be considered. There are 6 multiples of 8 less than 55 and so it should be 12..
Last two digit means remainder when devided by 100..
a) 101*102*103*197*198*199 / 100
=(01*99)*(02*9*(03*97) /100
now use negative remainder
=(- 1)*(-4)*(-6)/100
=(-24)/100
so remainder is 100 - 24 = 76..
*(03*97) /100
(03*97) gives (03*-3)ie..-9..can't be -6
(-1)*(-4)*(-9)/100
(-36)/100=64