Yeah..Nw i c.. 
@maroof10 said:Is n^2+3n+5 divisible by 121?....plz share the approach
It would never be divisible by 121..
for n^2+3n+5 div by 121, it would also be div by 11
or n^2-8n+16+11n-11 would be div by 11
or (n-4)^2+11(n-1) must be div by 11
so (n-4) = 11k
=> n = 11k+4
put it in the child eq
(11k+4)^2+3(11k+4)+5 = 121k^2+88k+16+33k+12+5 = 121k^2+121k+33 = 121k(k+1)+33
which would always give 33 as remainder on division with 121....hence, not divisible...
@maroof10
Now, n^2 + 3n + 5 => n^2 - 8n + 16 + 11n - 11 => (n-4)^2 - 11(n-1)
For the abve Eqn to be divisible by 121, it shld be divisible by 11..Thus, conversely, (n-4)^2 shld be divisible by 11...
Now, this is only possible when n-4 =11k..where k is sum integer..
Thus, n = 11k + 4..
Now, n^2 + 3n + 5=> 121k^2 + 121k + 33 => 121k(k+1) + 33
The above Eqn will always give a remainder of 33 when divided by 121..Hence it will never be divisible by 121..
1> last two digits of 1/5^903
2> 10000! = (100!)^k * P , P and are Integers What can be the max value of K ?
105
102
103
104
@joyjitpal said:approach?
How many integer values of x & y satisfy the expression 4x+7y=3 where lxl
4x + 7y = 3
=> x = (3-7y)/4
CASE 1: when y is a +ve number from [0, 1000) (note open brackets for 1000 as per question)
There are 4 possibilities for y = 4k or 4k+1 or 4k+2 or 4k+3 (ie multiple of 4 or others)
Now
(a) if y =4k
x = [3 - 7(4k) ] /4 = 3/4 - 7k => x not integer
(b) if y = 4k+1
x= -1- 7k => x is an integer
similarly other cases would give that x is NOT an integer.
So possible values for y are all the positive numbers of the form (4k+1) = 5,9,...,571
A total of 142
CASE 2 : when y is a -ve number from [0, -1000) (note open brackets for -1000 as per question)
Consider same technique and the possible form in this case would be (4k+3)
ie -3, -7, -11, ...., -567 (NOTE that 571 gives x=1000 but question says
A total of 142
Hence answer is 142+142 = 284 (not 285 as i said in my last post)
Hope the answer is right. Can anyone verify ??
@Shrutim90 said:How many words can be formed of the letters in the word RAINBOW, so that the vowels may occupy only even positions?I solved it in this way: R_N_B_W_These four positions are even positions which can be occupied by 3 vowels.So Ans is 4P3 * 4!But the solution given in the book is:_A_I_O_4! * 3!While I have no beef with the solution given in the book, I don't understand why my approach is incorrect !
We can't have 4 possible spaces...only 3...If you take as you have and put the vowels in these positions RXNXB_WX then they are 2,4 and 7 position! You can't leave a blank space in between in the middle of a word yaar...
regards
scrabbler
regards
scrabbler
@amresh_maverick said:1> last two digits of 1/5^903
2^903/10^903
(2^10)^90*2^3*10^-903
24^90*2^3*10^-903
76^45*2^3*10^-903
76*08*10^-903
08*10^-903
so 08
@amresh_maverick said:1> last two digits of 1/5^9032> 10000! = (100!)^k * P , P and are Integers What can be the max value of K ?105102103104
1> 08.
(1/5)^903 = (2/10)^903
or 0.2^903
or for last two digits.. find last two digits of 2^903 ... =2^20*2^883= 76*2^3=08.
(since last two digits of 2^20 = 76)
(1/5)^903 = (2/10)^903
or 0.2^903
or for last two digits.. find last two digits of 2^903 ... =2^20*2^883= 76*2^3=08.
(since last two digits of 2^20 = 76)
@amresh_maverick said:1> last two digits of 1/5^9032> 10000! = (100!)^k * P , P and are Integers What can be the max value of K ?105102103104
1)08
2)103
@amresh_maverick said:1> last two digits of 1/5^9032> 10000! = (100!)^k * P , P and are Integers What can be the max value of K ?105102103104
1> 08.
(1/5)^903 = (2/10)^903
or 0.2^903
or for last two digits.. find last two digits of 2^903 ... =2^20*2^883= 76*2^3=08.
(since last two digits of 2^20 = 76)
(1/5)^903 = (2/10)^903
or 0.2^903
or for last two digits.. find last two digits of 2^903 ... =2^20*2^883= 76*2^3=08.
(since last two digits of 2^20 = 76)
@surajmenonv said:@MANJULNEOGIHow many integer values of x & y satisfy the expression 4x+7y=3 where lxl 4x + 7y = 3 => x = (3-7y)/4CASE 1: when y is a +ve number from [0, 1000) (note open brackets for 1000 as per question)There are 4 possibilities for y = 4k or 4k+1 or 4k+2 or 4k+3 (ie multiple of 4 or others)Now (a) if y =4k x = [3 - 7(4k) ] /4 = 3/4 - 7k => x not integer (b) if y = 4k+1 x= -1- 7k => x is an integer similarly other cases would give that x is NOT an integer.So possible values for y are all the positive numbers of the form (4k+1) = 5,9,...,571A total of 142CASE 2 : when y is a -ve number from [0, -1000) (note open brackets for -1000 as per question)Consider same technique and the possible form in this case would be (4k+3)ie -3, -7, -11, ...., -567 (NOTE that 571 gives x=1000 but question says A total of 142Hence answer is 142+142 = 284 (not 285 as i said in my last post)Hope the answer is right. Can anyone verify ??@joyjitpal @MANJULNEOGI
dude it must be 285.. bohat simple hai.. one the positive number line the A.P starts from 6 and not 1.. so the total number of terms should be 142.. on the negative side it starts from -1 so there should be 143
To make this question simpler all you have to do is to find out which variable has a larger common difference span.. in this case it is x as it wil have a common difference of 7.. and also the question is |x|
To make this question simpler all you have to do is to find out which variable has a larger common difference span.. in this case it is x as it wil have a common difference of 7.. and also the question is |x|
@surajmenonv said:@MANJULNEOGIHow many integer values of x & y satisfy the expression 4x+7y=3 where lxl 4x + 7y = 3 => x = (3-7y)/4CASE 1: when y is a +ve number from [0, 1000) (note open brackets for 1000 as per question)There are 4 possibilities for y = 4k or 4k+1 or 4k+2 or 4k+3 (ie multiple of 4 or others)Now (a) if y =4k x = [3 - 7(4k) ] /4 = 3/4 - 7k => x not integer (b) if y = 4k+1 x= -1- 7k => x is an integer similarly other cases would give that x is NOT an integer.So possible values for y are all the positive numbers of the form (4k+1) = 5,9,...,571
Don't forget k can be 0 and so y = 1 bhi count karna padega.
@surajmenonv said:A total of 142CASE 2 : when y is a -ve number from [0, -1000) (note open brackets for -1000 as per question)Consider same technique and the possible form in this case would be (4k+3)ie -3, -7, -11, ...., -567 (NOTE that 571 gives x=1000 but question says A total of 142Hence answer is 142+142 = 284 (not 285 as i said in my last post)
Hence the 285 was correct :)
@ChirpiBird said:142?
Mod hai so negative waale cases bhi include karo!
regards
scrabbler
regards
scrabbler
@amresh_maverick said:1> last two digits of 1/5^9032> 10000! = (100!)^k * P , P and are Integers What can be the max value of K ?105102103104
1)Is it 08??
1/5^903 can be re-written as 2^903/10^903...now just find the last 2 digits of 2^903
=> 2^3*(2^10)^90 = 8*76 = 08..
2) Is it 103??
P = 10000!/(100!)^k
for P to be an integer, (100!)^k must perfectly divide 10000!
number of zeros in 10000! = 10000/5 = 2000/5 = 400/5 = 80/5 = 16/5 = 3
=> 2000+400+80+16+3 = 2499
number od zeros in 100! = 100/5 = 20/5 = 4
=> 20+4 = 24
for max k, p must be 1..
=> max(k) = [2499/24] = 104...
but let us also count the number of 2's
number of 2's in 10000! = 10000/2 = 5000/2 = 2500/2 = 1250/2 = 625/2 = 312/2 = 156/2 = 78/2 = 39/2 = 19/2 = 9/2 = 4/2 = 2/2 = 1
=> 5000+2500+1250+625+312+156+78+39+19+9+4+2+1 = 9995
number of 2's in 100! = 97
hence, max(k) = [9995/9] = 103...
so it should be 103...
@surajmenonv said:@MANJULNEOGIHow many integer values of x & y satisfy the expression 4x+7y=3 where lxl 4x + 7y = 3 => x = (3-7y)/4CASE 1: when y is a +ve number from [0, 1000) (note open brackets for 1000 as per question)There are 4 possibilities for y = 4k or 4k+1 or 4k+2 or 4k+3 (ie multiple of 4 or others)Now (a) if y =4k x = [3 - 7(4k) ] /4 = 3/4 - 7k => x not integer (b) if y = 4k+1 x= -1- 7k => x is an integer similarly other cases would give that x is NOT an integer.So possible values for y are all the positive numbers of the form (4k+1) = 5,9,...,571A total of 142CASE 2 : when y is a -ve number from [0, -1000) (note open brackets for -1000 as per question)Consider same technique and the possible form in this case would be (4k+3)ie -3, -7, -11, ...., -567 (NOTE that 571 gives x=1000 but question says A total of 142Hence answer is 142+142 = 284 (not 285 as i said in my last post)Hope the answer is right. Can anyone verify ??@joyjitpal @MANJULNEOGI
.. did exactly this way
got 142 only..
just forgot to add both cases. lol.
even i think 284 is right.
got 142 only..
just forgot to add both cases. lol.
even i think 284 is right.
The capitals of 4 partners A B C D are in the ratio of 7:8:6:5 . A and C's capitals are there in the business for the entire year.If each partner kept his money invested in the business for a period which is more than 6 months , and B and D together get 111/267 of the total profit,then for how many months is D's capital invested?
a)11
b)7
c)9
d)cant be determined
@scrabbler said:Don't forget k can be 0 and so y = 1 bhi count karna padega.Hence the 285 was correct
hehe my bad :p
@ChirpiBird said:.. did exactly this way got 142 only.. just forgot to add both cases. lol. even i think 284 is right.
285 hai 😛 first case me k=0 bhool gaya :P
A 25 meter long wound cable is cut into 2 and 3 meter long pieces. How many different ways can this be done if also the order of pieces of different lengths is taken into account?
@iLoveTorres said:also the question is |x|
In certain cases it can be 286, not 284...2001/7 = 285.71 so either 285 or 286 hoga...sorry for nit-picking :)
regards
scrabbler
regards
scrabbler