lod 2 ques.No: 79
Aset is formed by including some of the first one tousand natural numbers. S contains the maximum number of numbers such that they satisfy the following conditions:
1. No number of the set S is prime.
2.when the numbers of the set are selected 2 at a time, we always see co prime numbers.
The answer i got is 11.
the set contains the squares of prime numbers from 2.......31.
4,9,25,49.121,13^2,17^2,19^2,23^2,29^2,31^2. is the answer i got
am i missing any number here(because in the book the answer is given as 12.)
checkout and clear my doubt uys....
thanks vidhyasagar
lod 2 question no: 78
what is the remainder when (1!)^3+(2!)^3+(3!)^3+.......+(1152!)^3 is divided by 1152?
a)125 b)225 c)325 d)205
please explain the answwer i n detail guys....
Thanks
vidhyasagar
lod 2 question no: 78
what is the remainder when (1!)^3+(2!)^3+(3!)^3+.......+(1152!)^3 is divided by 1152?
a)125 b)225 c)325 d)205
please explain the answwer i n detail guys....
Thanks
vidhyasagar
1152=3^2*2^7
now (4!)^3 onwards is divisible by 1152. now we need to check with nos. before that.
(1!)^3+(2!)^3+(3!)^3=1+8+216=225..
answer is option 'b'
1152=3^2*2^7
now (4!)^3 onwards is divisible by 1152. now we need to check with nos. before that.
(1!)^3+(2!)^3+(3!)^3=1+8+216=225..
answer is option 'b'
Thanks a lot man....
You really rock...
Keep it up..
Regards
vidhyasagar
Lod 2 Q.no 88
Arun, bikas and chetakar have a total f 80 coins among them. arun triples the number of coins with the others by giving them some coins from his whole colection. next, bikas repeats the same process. after this bikas now has 20 coins. find the number of coins he had at the beginning?
a)11 b)10 c)9 d)12
Please explain the answer guys....
thanks in advance
cheers
vidhyasagar
Lod 2 Q.no 88
Arun, bikas and chetakar have a total f 80 coins among them. arun triples the number of coins with the others by giving them some coins from his whole colection. next, bikas repeats the same process. after this bikas now has 20 coins. find the number of coins he had at the beginning?
a)11 b)10 c)9 d)12
Please explain the answer guys....
thanks in advance
cheers
vidhyasagar
I am getting 20 as answer But its not there in the option :sleepy:
bikas --- -Chetan --|--Arun
20 ------ x ------ 60-x
20+2x/3+40-2x/3=60 ------ x/3 ------(60-x)/3
20 ----|---x/9------- 60-x/9
I am getting 20 as answer But its not there in the option :sleepy:
bikas --- -Chetan --|--Arun
20 ------ x ------ 60-x
20+2x/3+40-2x/3=60 ------ x/3 ------(60-x)/3
20 ----|---x/9------- 60-x/9
I am also getting 20 which is not in the option.
This is how i did it.
Arun - a coins
Bikas - b coins
Chetakar - c coins
After first step
Arun = a-2b-2c
Bikas= 3b
Chetakar=3c
After second step
Bikas has = 3b - 2(a-2b-2c) -2(3c)
=7b-2a-2c
= 9b -2(a+b+c)
20 =9b -2(80)
b=20
lod2-numbers
Define number k such that sum of squares of 1st M natural nos.(ie k=1^2+2^2......+M^2) when Ma)10b)11c)12d)none
can anyone...solve dis 4 me...
lod2-numbers
Define number k such that sum of squares of 1st M natural nos.(ie k=1^2+2^2......+M^2) when Ma)10b)11c)12d)none
can anyone...solve dis 4 me...
the formula for sum of squares of M consecutive integers = M*(M+1)*(2M+1)/6 = A.
now 2M+1 will always be odd.
CASE 1:
whn M is divisible by 3 and odd, (M+1) will be even. whn (m+1) is reduced with 2, the quotient obtained shud be furteher divisible by 4.
this is only possible if (M+1) is a multiple of 8.
CASE 2:
whn m is even, using similar reasoning it shud be divisible by 8.
so the no of possible solutions
= *2=6*2=12 [] denotes gr8st integer function
ans is option 'c'
1152=3^2*2^7
now (4!)^3 onwards is divisible by 1152. now we need to check with nos. before that.
(1!)^3+(2!)^3+(3!)^3=1+8+216=225..
answer is option 'b'
dude..can u explain the highlightd part....how can u say 4!^3 onward is divisible by 1152?????
the formula for sum of squares of M consecutive integers = M*(M+1)*(2M+1)
here M or (M+1) can be a multiple of 4.
so total no of values= *2= 26
where []denotes gr8st integer function
ans is option 4
but yaar answer in the book is given as c)12...:|
ashwinrulz
sorry my post was wrong....post edited....extremely sorry for the mistake..
was really sleepy when i solved n skipped the divided by 6 part...
my apologies
hey can any1 help me out with the logic behind findin last two digits in an question with examples.......
and help me out with des problmz...
1)101*102*103*197*198*199
2)(201*202*203*204*246*247*248*249)^2
hey can any1 help me out with the logic behind findin last two digits in an question with examples.......
and help me out with des problmz...
1)101*102*103*197*198*199
2)(201*202*203*204*246*247*248*249)^2
when u have to find out the last two digits its same as finding out the remainder when divided by 100.
1) remainder wen divided by 100 = 2*3*(-3)*(-2)*(-1) =100-36 = 64.
2) remainder = (2*3*4*46*47*48*49)^2
= ^2
= ^2
= ^2
= (44*46)^2
= (24)^2
= 76
lod2-numbers
Define number k such that sum of squares of 1st M natural nos.(ie k=1^2+2^2......+M^2) when Ma)10b)11c)12d)none
can anyone...solve dis 4 me...
n(n+1)(2n+1) must be a multiple of 24.
2n+1 and one out of n or n+1 are odds.Hence either n or n+1 must be a multiple of 8.
Also, n(n+1)(2n+1) by 3=>-(n-1)n(n+1) which is always a multiple of 6.
Hence the number of values of M = 2*=12 values
You can verify that by checking on the values= 7,8,15,16,23,24,31,32,39,40,47,48.
Hope your querry is answered:)
when u have to find out the last two digits its same as finding out the remainder when divided by 100.
1) remainder wen divided by 100 = 2*3*(-3)*(-2)*(-1) =100-36 = 64.
2) remainder = (2*3*4*46*47*48*49)^2
= ^2
= ^2
= ^2
= (44*46)^2
= (24)^2
= 76
dude....its fine with such nos...but wht if the question is smethin like dis
65*29*37*63*71*87*85 .find last 2 digits....
here if v go by remainder mehod..wnt help us in reducin the calculaion...
how shud v tacle such ones?????
when u have to find out the last two digits its same as finding out the remainder when divided by 100.
1) remainder wen divided by 100 = 2*3*(-3)*(-2)*(-1) =100-36 = 64.
2) remainder = (2*3*4*46*47*48*49)^2
= ^2
= ^2
= ^2
= (44*46)^2
= (24)^2
= 76
an u exlain the highlightd part?????
it shuld b ^2
thn hw have u proceeded???
puyz help me out in des problemz...
numbers-lod2
89)The computer takes input of a no. N & a X where X is a factor of no. N.In particular case N is equal to 83p796161q and X is equal to 11 where 0
Can anyone help me out with these?
1. Define a number K such that it is the sum of the squares of the first M natural numbers (k = 1^2 +2^2 + 3^2 .....+M^2) where M
a) 10 b)11
c)12 d)none of these
2. Find the 28383rd term of series 123456789101112....
a)3 b)4
c)9 d)7
First one I don't have a clue, don't know how to proceed further from K=m(m+1)(2m+1)/6.
And the second one I am getting answer 3 whereas books says 9.
1. Define a number K such that it is the sum of the squares of the first M natural numbers (k = 1^2 +2^2 + 3^2 .....+M^2) where M
a) 10 b)11
c)12 d)none of these
2. Find the 28383rd term of series 123456789101112....
a)3 b)4
c)9 d)7
First one I don't have a clue, don't know how to proceed further from K=m(m+1)(2m+1)/6.
And the second one I am getting answer 3 whereas books says 9.
puyz help me out in des problemz...
numbers-lod2
89)The computer takes input of a no. N & a X where X is a factor of no. N.In particular case N is equal to 83p796161q and X is equal to 11 where 0
Answer of 92.
14. 7! + 14 .13!
2.8! -21.6!
= 14(7! +13!)
6!(2.8.7 -21)
= 14( 7!+13!)
91(6!)
=14(7!+13!)
7!(13)
= 14.7! + 14.13!
7!.13 7!.13
= rem 7! + rem 0
= rem 7 !
ans