@jain4444 said:Find an 8-digit number using 4 2's and 4 1's that is divisible by 3 and 16.
_ _ _ _ 2112 -> 4!/2!2!=6 numbers how can it be 1 number?
@hedonistajay said:Find the remainder (6^83+8^83) is divided by 49?Options: 1 , 0 ,25, 35PS : do not have OA so kindly give the approach
E(49) equal to 42
(6^84+8^84) divided by 49 remainder is 2
but (6^83+8^83) ye kaise nikalna hai
@jain4444 plz help bhai :)
(6^84+8^84) divided by 49 remainder is 2
but (6^83+8^83) ye kaise nikalna hai
@jain4444 plz help bhai :)
@joyjitpal said:E(49) equal to 42(6^84+8^84) divided by 49 remainder is 2but (6^83+8^83) ye kaise nikalna hai@jain4444 plz help bhai
reverse eluer laga do bhai
6^83 * 6 mod 49 = 1
let 6^83 mod 49 = R
let 6^83 mod 49 = R
6R mod 49 = 1
=> R = 41
same for 8^83
@jain4444 said:reverse eluer laga do bhai 6^83 * 6 mod 49 = 1let 6^83 mod 49 = R 6R mod 49 = 1 => R = 41 same for 8^83
6R mod 49 = 1=> R = 41 could u pls explain how . Shud we keep seeing the multiples of 49 to the point where we wud get 1, or is there an easier way ?
@jain4444 said:reverse eluer laga do bhai 6^83 * 6 mod 49 = 1let 6^83 mod 49 = R 6R mod 49 = 1 => R = 41 same for 8^83
i could not get the bold part :(
@Calvin4ever said:6R mod 49 = 1=> R = 41 could u pls explain how . Shud we keep seeing the multiples of 49 to the point where we wud get 1, or is there an easier way ?
6R mod 49 = 1 can also be written as
-(-6R) mod 49 = 1 = OR (-6)(R) mod 49 = -1 = 48
From here, you can see: R mod 49 = -8
Hence, R =41
@Calvin4ever said:6R mod 49 = 1=> R = 41 could u pls explain how . Shud we keep seeing the multiples of 49 to the point where we wud get 1, or is there an easier way ?
@joyjitpal said:i could not get the bold part
this is nothing but chinese remainder theorem
6R = 49k + 1 --------- (1)
49 mod 6 = 1
6R = 1k + 1
=> put = k = 5
not put value of k in eq. (1)
6R = 246
=> R = 41
by some practice you will be able do it easily
@ScareCrow28 said:6R mod 49 = 1 can also be written as -(-6R) mod 49 = 1 = OR (-6)(R) mod 49 = -1 = 48 From here, you can see: R mod 49 = -8Hence, R =41
got it thanxx :)
find the number of numbers such that when divided by 10 it is a perfect square, when divided by 12 it is a perfect cube and when multiplied by 5/2 it is perfect power of 5.
@jain4444 said:
find the number of numbers such that when divided by 10 it is a perfect square, when divided by 12 it is a perfect cube and when multiplied by 5/2 it is perfect power of 5.
0 hai kya?
No is of the form = N = 10*(5^a) which is not divisible by 3 ..
@jain4444 said:find the number of numbers such that when divided by 10 it is a perfect square, when divided by 12 it is a perfect cube and when multiplied by 5/2 it is perfect power of 5.
Not possible !!?
The sum of the n terms of a series is n! + n^2 then the 6th term is . . ?? (n is a natural number)
@jain4444 said:find the number of numbers such that when divided by 10 it is a perfect square, when divided by 12 it is a perfect cube and when multiplied by 5/2 it is perfect power of 5.
Bhai it seems its not possible...
But m not sure...hit and trail kiya hai by seeing various possibilities....
OA kya hai??
@ScareCrow28 said:0 hai kya?
@Calvin4ever said:Not possible !!?
smallest will be = 2^11 * 3^10 * 5^9
and number of numbers will be infinite like 2^11 * 3^10 * 5^9 * 7^30 * 11^30 and so on
@Calvin4ever said:The sum of the n terms of a series is n! + n^2 then the 6th term is . . ?? (n is a natural number)
options nai hain kya???
@Calvin4ever said:The sum of the n terms of a series is n! + n^2 then the 6th term is . . ?? (n is a natural number)
6! + 6^2 - 5! - 5^2 = 600 + 11 = 611 ?
@Calvin4ever said:The sum of the n terms of a series is n! + n^2 then the 6th term is . . ?? (n is a natural number)
21...
@jain4444 said:smallest will be = 2^11 * 3^10 * 5^9and number of numbers will be infinite like 2^11 * 3^10 * 5^9 * 7^30 * 11^30 and so on
perfect power of 5 is x^5 and I took it as 5^x !! 
@jain4444 said:smallest will be = 2^11 * 3^10 * 5^9and number of numbers will be infinite like 2^11 * 3^10 * 5^9 * 7^30 * 11^30 and so on
Perfect power of 5 kaha hai when multiplied by 5/2 ?? 3 to agaya na isme?? I couldn't understand the meaning perfect power of 5. I thought it contains powers of 5 only
I couldn't understand the meaning perfect power of 5. I thought it contains powers of 5 only