its OA is D
its OA is D
What is the remainder when the number 123123123.....123123 (300 digits) is divided by 99?
a) 18 b) 27 c) 33 d) 36
a) 18 b) 27 c) 33 d) 36
@abhishekj1991 said:find a remainder of 2^133 divided by 133?
128?
E(133) = 108
2^108 mod 133 =
2^25 mod 133 = 2^4*2^21 mod 133 = 16*(125)^3 mod 133 = 16*(-5)^3 mod 133 = 128
@joyjitpal said:What is the remainder when the number 123123123.....123123 (300 digits) is divided by 99?a) 18 b) 27 c) 33 d) 36
33
@abhishekj1991 said:@sbharadwaj how?
Use chinese reminder theorem or Eulers.
I used chinese..
2^133 / 133
Bifurcate 133 as 2 co-primes (or primes as well )
2^133/ 7*19
Solve them separately.,
2^133 % 7 = a (say) and 2^133 % 19 = b
Now,
7x + a = 19y + b.
Solve the equation for least possible number. That'll be your reminder 😃
Hope this helps..!!
@raopradeep said:what is the minimum value of the function |x-2|+|x-1|+|x+11| where x is real numbera)11B)13c)12d)14
13?
at x =1
Find the remainder when (123123123123..................300digits) is divided by 37.
@ajeetaryans said:133 is a prime noand according to fermat theorema^p /p where P is a prime no remainder will be aso the answer is 2
sorry friend my mistake
if 133 = 19 * 7
then
first divide by 7
2 ^133 /7
= { 2 * 2 ^ 132 } /7
= { 2 * 8^44 } /7
= remainder is 2 when divide by 7
when divide by 19
reminder will be
2 ^ 133 /19 ->
- > (2^19)^7 / 19
-> 2^7 /19
128 /19 = 14 remainder
so now the answer
least no when divide by 7 gives remainder 2 and divide by 19 gives remainder 14 will be
7 m + 2 = 19 n + 14
m = { 19 n + 14 -2 } / 7
= { 19 n +12 } / 7
= 5 n + 5 / 7 (so put value of n such that it should be complete divisible )
n = 6 it will be complete divisible
so the remainder
= 19 n + 14
= 114 + 14
= 128
128 will be the remainder
if 133 = 19 * 7
then
first divide by 7
2 ^133 /7
= { 2 * 2 ^ 132 } /7
= { 2 * 8^44 } /7
= remainder is 2 when divide by 7
when divide by 19
reminder will be
2 ^ 133 /19 ->
- > (2^19)^7 / 19
-> 2^7 /19
128 /19 = 14 remainder
so now the answer
least no when divide by 7 gives remainder 2 and divide by 19 gives remainder 14 will be
7 m + 2 = 19 n + 14
m = { 19 n + 14 -2 } / 7
= { 19 n +12 } / 7
= 5 n + 5 / 7 (so put value of n such that it should be complete divisible )
n = 6 it will be complete divisible
so the remainder
= 19 n + 14
= 114 + 14
= 128
128 will be the remainder
@abhishekj1991 said:Find the remainder when (123123123123..................300digits) is divided by 37.
16
@raopradeep said:what is the minimum value of the function |x-2|+|x-1|+|x+11| where x is real numbera)11B)13c)12d)14
13
@joyjitpal said:What is the remainder when the number 123123123.....123123 (300 digits) is divided by 99?a) 18 b) 27 c) 33 d) 36
123123.................(300 digits)
=123(1+10^3+10^6+10^9+....................10^297)
taking modulo 99, we get
24*(1+10+1+10+1..........................+10)
24*550mod99
=24*55mod99
=33
@raopradeep said:what is the minimum value of the function |x-2|+|x-1|+|x+11| where x is real numbera)11B)13c)12d)14
13
13
@abhishekj1991 said:Find the remainder when (123123123123..................300digits) is divided by 37.
37 * 3 = 111
111 * 9 = 999
so take an small example like
123123
123 * 1000 + 123
= { 123 (999 +1) + 123 } / 37
so remainder wil be {123 *2 } / 37
in same manner
when 300 digits total pairs are 100
( 123 * 100) / 37
= 12 * -11 /37
= -132 /37
= -21
= 16
so remainder will be 16
@bs0409 said:123123.................(300 digits)=123(1+10^3+10^6+10^9+....................10^297)taking modulo 99, we get24*(1+10+1+10+1..........................+10)24*550mod99=24*55mod99=33
bro, can u explain d part in bold above?
take 123 common and then 10^297+................+ 10^0
@bs0409 bhai has written in different way ie. from10^0 to 10^297.