hi puys ,
pls help me out in this question :
2^1002 / 1003 what will b d remainder ???
please explain with solution !!!
I think It should be 1
Use Fermat's little theorem:For any number N not divisible by p, N^(p-1) = 1 mod p where p is a prime
I think It should be 1
Use Fermat's little theorem:For any number N not divisible by p, N^(p-1) = 1 mod p where p is a prime
can anyone solve d following question:-
(6^83+8^83)/49 remainder will be???
i solved n gt the answer as 14 but my answer is wrong....can u plz solve
Should be 35.
E(49) = 42.
6^84 mod 49 = 1.
=> 6*6^83 mod 49 = 1.
=> 6*X mod 49 = 1.
=> X = 41.
8^84 mod 49 = 1.
=> 8*Y mod 49 = 1.
=> Y = 43.
So, (X+Y) mod 49 = (41+43) mod 49 = (-8-6) mod 49 = -14 or 35.
can anyone solve d following question:-
(6^83+8^83)/49 remainder will be???
i solved n gt the answer as 14 but my answer is wrong....can u plz solve
(7-1)^83 + (7+1)^83 mod 49
on binomial expansion the terms that will be left are
7-83+7+83=14 mod 49 = 35
Should be 35.
-14 or 35.
(7-1)^83 + (7+1)^83 mod 49
on binomial expansion the terms that will be left are
7-83+7+83=14 mod 49 = 35
Can you guyz show me where I am wrong:-
The last exp. that I got was 14/49........after that ?
PS:- Bino. Exp. method plzz..
Regards,
Never Back Down(7-1)^83 + (7+1)^83 mod 49
on binomial expansion the terms that will be left are
7-83+7+83=14 mod 49 = 35
Yar how can 14 mod 49 be 35?
14 mod 49 has to be 14 it self.
Pls check.
I think It should be 1
Use Fermat's little theorem:For any number N not divisible by p, N^(p-1) = 1 mod p where p is a prime
1003 is not a prime number !!!!!!
chitrartha Says1003 is not a prime number !!!!!!
2 and 1003 are reletively prime to each other so we can apply the above said theorm in the question.
When 2 zeros, three possibilities:-
i) a00b
b - a should be divisible by 7
when b = a, 9 cases
b - a = 7, 2 cases
a - b = 7, 2 cases
13 cases
ii) ab00
ab should be a multiple of 7
=> 13 such numbers
iii) a0b0
=> 3b - a should be be divisible by 7 (a should not be 0)
=> 3b = 7k + a
=> (a, b) can be (1, 5), (2, 3), (3, 1), (3,, (4, 6), (5, 4), (6, 2), (6, 9), (7 7), (8, 5)
=> 10 such numbers
When 3 zeros, one possibility
a000
=> one such number
=> Total = 37 such numbers
, (4, 6), (5, 4), (6, 2), (6, 9), (7 7), (8, 5)Please explain the 3rd part
Puys please help me solve this ......
109^53
what are the last two digits?
please explain me how to solve these questions......
Can anyone tell me the formula to calculate E(summation)
Puys please help me solve this ......
109^53
what are the last two digits?
please explain me how to solve these questions......
For last two digits take the remainder with 100
....
Puys please help me solve this ......
109^53
what are the last two digits?
please explain me how to solve these questions......
Would be 29.
For last two digits just take remainder with 100.
109^53 mod 100 = 9^53 mod 100.
E(100) = 40.
=> 9^13 mod 100 = 9*9^12 mod 100 = 9*(9^3)^4 mod 100 = 9*29^4 mod 100 = 9*41^2 mod 100 = 9*27 mod 100 = 729 mod 100 = 29.
Puys please help me solve this ......
109^53
what are the last two digits?
please explain me how to solve these questions......
Last two digits is nothing but the remainder by 100.
M Ankit SaysCan anyone tell me the formula to calculate E(summation)
In the simplest of manner,
E
= 1/2 E
= 1/2
Now u can use standard equations for summation.
whenever the base ends with an odd number.. try to get its unit digit as 1
in this case 109^53
we have
109(___81)^52
The last two digits of a1^bcd are given by
1) the last digit will be 1 always..
2) the second last digit will be the last digit of the product of a and d
so we have 109(___61)
so is it 49?
First of all thanks Insanatine for your prompt response.
I know the last digit will be 9 for sure........
But i dont know how to find out the ten's place.......
its not 49........:nono:
so i want to know how do we find out the ten's place ?
Last two digits is nothing but the remainder by 100.
In the simplest of manner,
E
= 1/2 E
= 1/2
Now u can use standard equations for summation.
Spectra sir on a QA (related) thread. ____/\____
Jivan safal ho gaya.
Enceladus , spectramind.. can you please point out the place i went wrong while finding out the last two digits? was doin it with the approach of getting the last digit as 1 and den solvin
Insanatine SaysEnceladus , spectramind.. can you please point out the place i went wrong while finding out the last two digits? was doin it with the approach of getting the last digit as 1 and den solvin
That can be done in the following manner.
109^53 = 09^53 . = (03)^106 = 3^2*(3^4)^26 = 9*81^26 = 9*81 (last two digits of 81^26 would be 81) = 729 or 29 as last two digits.
Hope it helps. 😃
Spectra sir on a QA (related) thread. ____/\____
Jivan safal ho gaya. :biggrin:
Bhai saab __/\__ dopahar tak pahunch jaunga Official thread pe.Fir dekhna lafandar approach
well i came through this question while solving today, since i am new to cat preparations a help from odrs might make a big difference to me..
n*(n^2 -1)*(5n+2) is always divisible by? please expalin your answers for n being both odd as well as even.
u can write it as, n*(n+1)*(n-1)*(5n+2)
(n+1)*(n-1) means whatever value of n you take, you will always have one odd and one even number in the expression, hence it will always be divisible by 6
Can anyone pls solve :
P Q R | S T
- 30 | - | 12 5
12 4 | 3 - -
Find P:Q:R:S:T?