@bs0409 said:If m is a natural no. and N = 2^m, what is the remainder when N! is divided by 2^N
2^m??
@ravi.theja said:Let 123abc231bca312cab be a 18 digit number in base 7. How many ordered pairs (a,b,c) exist such that the given 18 digit number is divisible by 4?
b+3a+c+2+1+1+a+3c+b+3+3+2+c+3b+a+1+2+3
5a+5b+5c+18
a+b+c+2
a+b+c=2,6,10,14,18
0a+b+c=2 =>4C2=6 ways
a+b+c=6=>8C2=28 ways
a+b+c=10 =>12C2-3*5C2=36 ways
a+b+c=14=>16C2-3*9C2+3C2=15 way
a+b+c=18=> 1way
5a+5b+5c+18
a+b+c+2
a+b+c=2,6,10,14,18
0a+b+c=2 =>4C2=6 ways
a+b+c=6=>8C2=28 ways
a+b+c=10 =>12C2-3*5C2=36 ways
a+b+c=14=>16C2-3*9C2+3C2=15 way
a+b+c=18=> 1way
86 ways
@gnehagarg said:b+3a+c+2+1+1+a+3c+b+3+3+2+c+3b+a+1+2+35a+5b+5c+19a+b+c+3a+b+c=1,5,9,13 ,170a+b+c=1 =>3C2=3 waysa+b+c=5=>7C2=21 waysa+b+c=9 =>11C2-3*4C2=37 waysa+b+c=13=>15C2-3*8C2=21 waya+b+c=17=> 19C2-3*12C2= -ve(can anyone explain what is error and that's why i got negative answer?)82 ways
OA: 86
@jaituteja said:@ravi.theja Yes... 7a + b divisible by 4.When a=0 , b =0,4,8a= 1 , b = 1,5,9a= 2, b = 2,6a= 3 , b=3,7a= 4, b = 0,4,8a= 5, b= 1,5,96 , b= 2,67, b= 3,78, b= 0,4,89, b = 1,5,910, b = 2,6..... Till a=14, b = 2 I think aise Honda cahaheyez.. Baki I m not sure..
bhai..the number is of base 7... a can take values from 0 to 6 only
@Budokai001 said:Rohan is asked to figure out the marks scored by Sunil in three different subjects with the help ofcertain clues. He is told that the product of the marks obtained by Sunil is 72 and the sum of themarks obtained by Sunil is equal to the Rohan €™s current age (in completed years). Rohan could notanswer the question with this information. When he was also told that Sunil got the highest marksin Physics among the three subjects, he immediately answered the question correctly. What is thesum of the marks scored by Sunil in the two subjects other than Physics?
3
@techgeek2050 said:@grkkrg can u plz explain d 1st solution... hcf( 17, 18) = 306 ??
It's not correct. (EDITED)
Two consecutive integers are always coprime to each other. So their LCM will be the product of the two numbers.
Two consecutive integers are always coprime to each other. So their LCM will be the product of the two numbers.
@grkkrg said:Two consecutive integers are always coprime to each other. So their HCF will be the product of the two numbers.
Not HCF. But it will be LCm which will be product of the two numbers.
@gnehagarg said:b+3a+c+2+1+1+a+3c+b+3+3+2+c+3b+a+1+2+3
5a+5b+5c+19
a+b+c+3
a+b+c=1,5,9,13 ,17
0a+b+c=1 =>3C2=3 ways
a+b+c=5=>7C2=21 ways
a+b+c=9 =>11C2-3*4C2=37 ways
a+b+c=13=>15C2-3*8C2=21 way
a+b+c=17=> 19C2-3*12C2= -ve(can anyone explain what is error and that's why i got negative answer?)
82 ways
for a+b+c=17, the answer shld be 19C2-3C1*12C2+3C2*5C2=3
The last expression is for cases where two of a,b,c are greater than 6.
A no. has exactly 32 factors out of which 4 are not composite.Product of these 4 factors (which are not composite) is 30.How many such no.s are possible???
@bs0409 said:A no. has exactly 32 factors out of which 4 are not composite.Product of these 4 factors (which are not composite) is 30.How many such no.s are possible???
6?
The number has 3 prime factors.
32 = 2^5
30 = 1 * 2 * 3 * 5
power of 2 = a
power of 3 = b
power of 5 = c
(a+1)(b+1)(c+1) = 2^5
1,1,7 => 3 ways
1,3,3 => 3 ways
The number has 3 prime factors.
32 = 2^5
30 = 1 * 2 * 3 * 5
power of 2 = a
power of 3 = b
power of 5 = c
(a+1)(b+1)(c+1) = 2^5
1,1,7 => 3 ways
1,3,3 => 3 ways
@bs0409 said:A no. has exactly 32 factors out of which 4 are not composite.Product of these 4 factors (which are not composite) is 30.How many such no.s are possible???
6??
@bs0409 said:If m is a natural no. and N = 2^m, what is the remainder when N! is divided by 2^N
N=2^m
N!=2^m!=2^(m-1)+2^(m-2)+2^(m-3)+2^(m-4)+----------------------------
2^(m-1)+2^(m-2)+2^(m-3)+---------------/(2^N)
Remainder is 0
@gnehagarg said:N=2^mN!=2^m!=2^(m-1)+2^(m-2)+2^(m-3)+2^(m-4)+----------------------------2^(m-1)+2^(m-2)+2^(m-3)+---------------/(2^N)Remainder is 0
ANS is not 0. Try again. I did not get your steps at all........
@bs0409 said:A no. has exactly 32 factors out of which 4 are not composite.Product of these 4 factors (which are not composite) is 30.How many such no.s are possible???
30=1*2*3*5
N=2^a*3^b*5^c
(a+1)(b+1)(c+1)=32 (2*2*8) (2*4*4)
2^1*3^1*5^7
2^1*3^3*5^3
If the sum of 101 distinct terms in arithmetic progression is zero, in how many ways can three if these terms be selected such that their sum is zero ?
@bs0409 said:A no. has exactly 32 factors out of which 4 are not composite.Product of these 4 factors (which are not composite) is 30.How many such no.s are possible???
30 = 15*2 = 3*5*2 = 1*2*3*5 (all not composite)
n = 2^x*3^y*5^z
32 = 2*2*2*2*2 = 2*4*4 = 2*2*8
2,2,4 = 3!/2! = 3 numbers
2,2,8 = 3!/2! = 3 numbers
hence total = 6
@bs0409 said:ANS is not 0. Try again. I did not get your steps at all........
I did 4=2^2 We have to calculate 4!.So,we do 4/2=2 and4/4=1.The total is 3
2^m!
2^m/2=2^(m-1)+2^(m-2)+---------------------------------infinity
The sum is 2^(m-1)
2^(m-1)/2^N
Remainder =0 (m-1>N)
Remainder=2^(m-1)(m-1
Is it wrong?