a. 1
By using fermat thm. u can hav it.
yes ans is 1
by euler's thm
The Question is:-
What is the Remainder when ^13333 is divided by 13?
6! and 13 are coprime
and here 7! has 12 in it so remainder will be 1 according to Euler's theoram
What is the Remainder when ^13333 is divided by 13?
Ans 1
6! = 720 7! = 7*720 (^13333)/13 = /13=(^13333)/13 = (^13333)/13 =(^1333)/13 = 1 so remainder = 1
Ans 1
6! = 720 7! = 7*720 (^13333)/13 = /13=(^13333)/13 = (^13333)/13 =(^1333)/13 = 1 so remainder = 1
What is the Remainder when ^13333 is divided by 13?
Ans 1
6! = 720 7! = 7*720 (^13333)/13 = /13=(^13333)/13 = (^13333)/13 =(^1333)/13 = 1 so remainder = 1
Ur method is long bro.... v cn do it easily using Euler's Theoram:-
It goes like this:-
as we know dt 13 is a prime number,So 6! is co-prime to it. And Euler number of 13 is 12 which is present in 7!. So this implies /13 . So according to the Euler's Theoram, Remainder comes out to be 1. 😁
6! and 13 are coprime
and here 7! has 12 in it so remainder will be 1 according to Euler's theoram
Yes, right. The answr is 1 n using Euler's Theoram/Fermet's Theoram 😁
Question:-
What will be the Unit Digit?
120
(2n-1)!
n=1
Question:-
What will be the Unit Digit?
120
(2n-1)!
n=1
my ans is 7 ...if its rite ill explain the approach
Question:-
What will be the Unit Digit?
120
(2n-1)!
n=1
my answer is 7...
since after 4! every factorial last digit is 0..
hence we only need to find..
last digit of 1! + 3!
=7..
prats92 Saysmy ans is 7 ...if its rite ill explain the approach
my answer is 7...
since after 4! every factorial last digit is 0..
hence we only need to find..
last digit of 1! + 3!
=7..
Yes the answer is Right and d Approach too...
Originally Posted by sid_roy09 View Post
Question:-
What will be the Unit Digit?
120
(2n-1)!
n=1
Ans 7 Series is = 1! + 3! + 5! +............................+239!
in this series every term starting from 5! will have ten's digit 0 so unit's digit will be contributed by 1! and 3! 1 + 3! = 1+6 = 7 so 7 is the unit's digit
1 for how many integer value of X is (2x^2-10x-4)/(x^2 -4x+3) is an integer
a) 4, b) 5 c) 6 d) 7
2. what is the largest prime number whose cube divides 1!2!3!......1001! ?
a) 997 b) 991 c) 977 d) 973
3. the highest power of 12 that can divide (5^36 - 1 ) is
a) 1 b) 2 c)3 d)4
2. what is the largest prime number whose cube divides 1!2!3!......1001! ?
a) 997 b) 991 c) 977 d) 973
I think its 997
Exponent of 997 in the given expression is 5, as it comes once in each of the last 5 factorials, i.e., 997!, 998!, 999!, 1000!, 1001!
3. the highest power of 12 that can divide (5^36 - 1 ) is
a) 1 b) 2 c)3 d)4
(5^36 - 1) = (5^18 - 1)(5^18 + 1) = (5^6 - 1)(5^12 + 5^6 + 1)(5^6 + 1)(5^12 - 5^6 + 1)
= (5^3 - 1)(5^3 + 1)(5^12 + 5^6 + 1)(5^2 + 1)(5^4 - 5^2 + 1)(5^12 - 5^6 + 1)
= (5 - 1)(5^2 + 5 + 1)(5 + 1)(5^2 - 5 + 1)(5^12 + 5^6 + 1)(5^2 + 1)(5^4 - 5^2 + 1)(5^12 - 5^6 + 1)
= 4*6(5^2 + 5 + 1)(5^2 - 5 + 1)(5^12 + 5^6 + 1)(5^2 + 1)(5^4 - 5^2 + 1)(5^12 - 5^6 + 1)
Now, all the brackets having three terms are not divisible by 2
SO, only terms left are 4*6(5^2 + 1)
=> Exponent of 2 is 4
=> Divisible by 16 ..................(1)
Now, for 3
6 is divisible by 3
5^2 - 5 + 1 is divisible by 3
5^12 + 5^6 + 1 is divisible by 3
=> Exponent of 3 is 3
=> Divisible by 27 ......................(2)
From (1) and (2), we can say that highest power of 12 is 2
Alternatively:-
Lets try to find the remainder when 5^36 is divided by 64. Eulers number for 64 is 32.
=> 5^36 = 5^4 (mod 64) = 125*5 (mod 64) = 49 (mod 64)
=> 5^36 - 1 = 48 (mod 64)
=> 5^36 - 1 = 64k + 48 = 16(4k + 3)
=> Highest exponent of 2 is 4, i.e., divisible by 16 (as in 12 exponent of 2 is 2 that means highest exponent of 12 can be atmost 2, so it will be enough to know whether the given number is divisible by 9 or not)
Lets try to find the remainder when 5^36 is divided by 9
Euler's number for 9 is 6
=> 5^36 = 1 (mod 9)
=> 5^36 - 1 = 0 (mod 9)
=> Its divisible by 9
=> Highest exponent of 12 is 2
Alternatively:-
Lets try to find the remainder when 5^36 is divided by 64. Eulers number for 64 is 32.
=> 5^36 = 5^4 (mod 64) = 125*5 (mod 64) = 49 (mod 64)
=> 5^36 - 1 = 48 (mod 64)
=> 5^36 - 1 = 64k + 48 = 16(4k + 3)
=> Highest exponent of 2 is 4, i.e., divisible by 16 (as in 12 exponent of 2 is 2 that means highest exponent of 12 can be atmost 2, so it will be enough to know whether the given number is divisible by 9 or not)
Lets try to find the remainder when 5^36 is divided by 9
Euler's number for 9 is 6
=> 5^36 = 1 (mod 9)
=> 5^36 - 1 = 0 (mod 9)
=> Its divisible by 9
=> Highest exponent of 12 is 2
I got the answer as 1
If we do it by plain binomial expansion,will the answer change???
1 for how many integer value of X is (2x^2-10x-4)/(x^2 -4x+3) is an integer
a) 4, b) 5 c) 6 d) 7
(2x^2-10x-4)/(x^2 -4x+3) = 2 - {(2x + 10)/{(x - 1)(x - 3)}}
For the above expression to be an integer, (2x + 10)/{(x - 1)(x - 3)} should be an integer.
When x is even, say 2k, then
(2x + 10)/{(x - 1)(x - 3)} = 2(2k + 5)/{(2k - 1)(2k - 3)}
Since (2k - 1) and (2k - 3) are co-prime to each other, both should divide 2k + 5
=> k = 1, 2
=> x = 2, 4
When x is odd, say 2k + 1, then
(2x + 10)/{(x - 1)(x - 3)} = (k + 3)/{k(k - 1)}
Since, k and k - 1 are co-prime to each other, both should divide k + 3
=> k = -3, -1, 3
=> x = -5, -1, 7
So, total 5 integral values
spectramind07 SaysI got the answer as 1If we do it by plain binomial expansion,will the answer change???
Using Binomial:-
For exponent of 2:-
5^36 - 1 = (4 + 1)^36 - 1
Lets consider last two terms
C(36, 2)*4^2 + C(36, 1)*4 = 32k + 144 = 16(2k + 9)
Since (2k + 9) is odd, highest power of 2 is 4
For exponent of 3
5^36 - 1 = (6 - 1)^36 - 1
Lets consider last three terms
-C(36, 3)*6^3 + C(36, 2)*36 - C(36, 1)*6
Since last term is divisible by 27
Highest exponent is 3
=> Highest exponent of 12 will be 2
Using Binomial:-
For exponent of 2:-
5^36 - 1 = (4 + 1)^36 - 1
Lets consider last two terms
C(36, 2)*4^2 + C(36, 1)*4 = 32k + 144 = 16(2k + 9)
Since (2k + 9) is odd, highest power of 2 is 4
For exponent of 3
5^36 - 1 = (6 - 1)^36 - 1
Lets consider last three terms
-C(36, 3)*6^3 + C(36, 2)*36 - C(36, 1)*6
Since last term is divisible by 27
Highest exponent is 3
=> Highest exponent of 12 will be 2
5^36 - 1
=(25)^18 - 1
=(24+1)^18 - 1
Continuing like this,we get the last term of the above expansion as (24)^1 (As the 1s cancel out).Taking 12 as common factor,we get 12(An integer).....So highest power i got is 12.Where am i going wrong??
5^36 - 1
=(25)^18 - 1
=(24+1)^18 - 1
Continuing like this,we get the last term of the above expansion as (24)^1 (As the 1s cancel out).Taking 12 as common factor,we get 12(An integer).....So highest power i got is 12.Where am i going wrong??
(24 + 1)^18 - 1
Lets consider last three terms
-C(18, 3)*24^3 + C(18, 2)*24^2 - 24*18
= -12^4k + 12^3n - 144*3
=> Highest power of 12 is 2
(24 + 1)^18 - 1
Lets consider last three terms
-C(18, 3)*24^3 + C(18, 2)*24^2 - 24*18
= -12^4k + 12^3n - 144*3
=> Highest power of 12 is 2
I think the thing you are missing is that in 12*integer the integer can itself be a multiple of 12.
Now after cancelling 1 we have , 24^18+k1*24^17+...................+18C2*24^2+18*24
Now the bold part is a multiple of 12^3 as 24^3=12^2*4 and 18C2 has two 3s.
So the term is 12^3*m+18*24
Now 18*24=12^2*3
So the number is 12^3*m+3*12^2 =12^2(12m+3)
So clearly divisible 12^2 only.
And that is what I also got though now I feel your method was an easier one
Thanks a lot.I was missing the 18 in the last but one term of the binomial expansion.
@ subha- binomial expansion is very useful at times sir.
2^3^4^5 - 2^3^5^4 ?
a) 0
b) 2
c) 4
d) 6
e) None of these
pls mention approach for soloving this type of question
(a^b^c)=(A)^(b*c)
e.g. 2^3^2=2^6=64.
that's why both terms 2^3^4^5 - 2^3^5^4 are same, hence zero.