3,12,24,33,66,?................next number in the series? guys plz tell the approach with ans.
@nagpal9 said:@sbharadwaj how...explain??
3,12,24,33,66
The series goes like
+9, x 2, +9, x2 and so on.
@nagpal9 said:3,12,24,33,66,?................next number in the series? guys plz tell the approach with ans.
check the sum
+ 9 , + 12 , + 9 , + 33 .. so next would be + 9
so 66 + 9 = 75.
+ 9 , + 12 , + 9 , + 33 .. so next would be + 9
so 66 + 9 = 75.
@nagpal9 said:3,12,24,33,66,?................next number in the series? guys plz tell the approach with ans.
you have to add 9 and multiply the numbers by 2 alternately. So add 9 to 3...That gives 12.
Now multiply 12 by 2 that gives 24..now again add 9 to 24...that gives 33...now multiply 33 by 2..that gives 66...now in last again add 9 to 66...that gives 75...hope it's clear...number after 75 will be 150 that is 75*2=150 and the next one will be 175+9=184..clear??
Now multiply 12 by 2 that gives 24..now again add 9 to 24...that gives 33...now multiply 33 by 2..that gives 66...now in last again add 9 to 66...that gives 75...hope it's clear...number after 75 will be 150 that is 75*2=150 and the next one will be 175+9=184..clear??
@ishu1991 said:@sunnychopra892^1000
i wonder where did all the prime nos go..
@sunnychopra89 said: 2000! / 1000 = N (1 X 3 X 5 ...... X 1997 X 1999). Find N?
it will be of the form 2^p * 3^q * 5*r * 7*s * .....
options would've made this easier.
OA?
@sunnychopra89 said: 2000! / 1000 = N (1 X 3 X 5 ...... X 1997 X 1999). Find N?
it will be of the form 2^p * 3^q * 5*r * 7*s * .....
options would've made this easier.
OA?
@x2maverickc said:i wonder where did all the prime nos go..
@sunnychopra89 said: 2000! / 1000 = N (1 X 3 X 5 ...... X 1997 X 1999). Find N?
it will be of the form 2^p * 3^q * 5*r * 7*s * .....
options would've made this easier.
OA?
Exatcly even I'm wondering....OA is 2^1000.....
@anuragu79
2000 ! = 1000n ( 1*3*5*7......1999)
2000! = (1*3*5..1999)(2*4*6*8...2000)
1000n = 2^1000 (1*2*3...1000)
1000n = 2^1000 * 1000!
n = 2^1000 * 999 ! ?
mine was wrng
2000 ! = 1000n ( 1*3*5*7......1999)
2000! = (1*3*5..1999)(2*4*6*8...2000)
1000n = 2^1000 (1*2*3...1000)
1000n = 2^1000 * 1000!
n = 2^1000 * 999 ! ?
mine was wrng
Easy one.... N= 16^3+17^3+18^3+19^3. Find remainder when it is divided by 70
@TootaHuaDil said:Easy one.... N= 16^3+17^3+18^3+19^3. Find remainder when it is divided by 70
36100 - 14400 mod 70 = 0
N = 55^3 + 17^3 - 72^3 is divisible by?
@TootaHuaDil said:Easy one.... N= 16^3+17^3+18^3+19^3. Find remainder when it is divided by 70
Nmod2=0
Nmod5=0
Nmod7=0
Nmod70=0
Nmod5=0
Nmod7=0
Nmod70=0
@grkkrg said:36100 - 14400 mod 70 = 0
ye approach to pta hi nahi tha...bhai ye kaise krte hain?? :O
@TootaHuaDil said:ye approach to pta hi nahi tha...bhai ye kaise krte hain??
sum to 19^3 - sum to 15^3 :mg: