antodaya Says2 and 1003 are reletively prime to each other so we can apply the above said theorm in the question.
According to fermat theorm : a^(n-1) / n = Rem (1)
where , n = prime no and
a & n should be co-prime
According to fermat theorm : a^(n-1) / n = Rem (1)
where , n = prime no and
a & n should be co-prime
hi puys ,
pls help me out in this question :
A is a set of 7 consecutive two digit numbers, none of these numbers being a multiple of 10. Reversing the numbers in set A forms numbers in set B. The difference between the sum of the elements in Set A and those in set B is 63.The smallest number in set A can be ? A)64 B)13 C)31 D)75
Pls explain with solution
hi puys ,
pls help me out in this question :
A is a set of 7 consecutive two digit numbers, none of these numbers being a multiple of 10. Reversing the numbers in set A forms numbers in set B. The difference between the sum of the elements in Set A and those in set B is 63.The smallest number in set A can be ? A)64 B)13 C)31 D)75
Pls explain with solution
.......................
hi puys ,
pls help me out in this question :
A is a set of 7 consecutive two digit numbers, none of these numbers being a multiple of 10. Reversing the numbers in set A forms numbers in set B. The difference between the sum of the elements in Set A and those in set B is 63.The smallest number in set A can be ? A)64 B)13 C)31 D)75
Pls explain with solution
Yar these type of questions can be easily cracked with options.
See, the questions says that no number is a multiple of 10. Hence 64 and 75 cannot be the smallest number in the set A.
So, we are left with 13 and 31. Looking at 13 clearly reflects that the difference between A and B will be way higher than 63. So it has to be 31. 😃
hello puys,
pls help me in solving these number series prblms..
1.8 9 13 12 8 9
12 A B C D E
The two series follow the same logic..wat will be C?
2.one number is wrong in the series given below..which one?
2 3 7 13 26 47 78
3.in the following question a series will be established if two of the numbers in quotes r interchanged...which numbers?
4 '0' '-7' '-45' '-20' '-94'
4. wat will be A?
14.8 17.2 A 22 2.8 41.2
1.8 9 13 12 8 9
12 A B C D E
The two series follow the same logic..wat will be C?
---------------------------------------------------------------
Ans is 16 ( Series is in the from of +1,+4,-1,-4 and so on)
------------------------------------------------------------------
2.one number is wrong in the series given below..which one?
2 3 7 13 26 47 78
---------------------------------------------------------------
Ans : 26 is wrong (not sure)
-------------------------------------------------------------
3.in the following question a series will be established if two of the numbers in quotes r interchanged...which numbers?
4 '0' '-7' '-45' '-20' '-94'
--------------------------------------------------------------
Ans : -45 and -20 should be replaced to get the series (series should be like n*2-8,n*2-7,n*2-6 and so on)
-----------------------------------------------------------------------------
4. wat will be A?
14.8 17.2 A 22 2.8 41.2
---------------------------------------------------------------------------
Ans : A should be 12.40 (the diffrences in series would be in 2.4,4.8,9.6,19.2 and so on.
---------------------------------------------------------------------------
Please confirm with the option..thnx...
Q1)Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4
a. 1
b. 2
c. 3
d. 4
e. more than 4
Many puys have posted answer for this question...can anybody illustrate the thought process while tackling such problem ??
how to find the remainder of 6^83/49
my way - euler (49) = 42
so 6^-1/49 : does anybody know how to deal with negative powers in finding remainder?
how to find the remainder of 6^83/49
my way - euler (49) = 42
so 6^-1/49 : does anybody know how to deal with negative powers in finding remainder?
You don't do it like this then.
As E(49) = 42.
=> 6^84 mod 49 = 1.
=> 6*6^83 mod 49 = 1.
=> 6X mod 49 = 1 .
Now, it gets easy as we need to find a number which when divided with 6 leaves a remainder 1 on division with 49.
Which would be 41.
Hope it is clear.
You don't do it like this then.
As E(49) = 42.
=> 6^84 mod 49 = 1.
=> 6*6^83 mod 49 = 1.
=> 6X mod 49 = 1 .
Now, it gets easy as we need to find a number which when divided with 6 leaves a remainder 1 on division with 49.
Which would be 41.
Hope it is clear.
6^83/49=(7-1)^83/49
now expand using binomial u will find that
terms beyond 7^2 is divisible by 49
hence take the terms of c(83,0)(-1)^83+c(83,0)*7^1*(-1)^82/49
on solving we get
-1+83*7/7^2 =580/49=41(remainder)
Kindly Help, 
N is the sum of the squares of three consecutive odd numbers such that all the digits of N are the same.. If N is a four digit number, then the value of N is ..
A. 3333
B. 5555
C. 7777
D. 9999
Thank you..
Kindly Help,
N is the sum of the squares of three consecutive odd numbers such that all the digits of N are the same.. If N is a four digit number, then the value of N is ..
A. 3333
B. 5555
C. 7777
D. 9999
Thank you..
let the three odd #are = n - 2 , n , n + 2
n^2 + 4 + n^2 + n^2 + 4
3n^2 + 8
now check the options by sub 8 and ishould be divisible by 3..only 5555 doin so...
Kindly Help,
N is the sum of the squares of three consecutive odd numbers such that all the digits of N are the same.. If N is a four digit number, then the value of N is ..
A. 3333
B. 5555
C. 7777
D. 9999
Thank you..
double post............
Kindly Help,
N is the sum of the squares of three consecutive odd numbers such that all the digits of N are the same.. If N is a four digit number, then the value of N is ..
A. 3333
B. 5555
C. 7777
D. 9999
Thank you..
is it b????
Kindly Help,
N is the sum of the squares of three consecutive odd numbers such that all the digits of N are the same.. If N is a four digit number, then the value of N is ..
A. 3333
B. 5555
C. 7777
D. 9999
Thank you..
Three consecutive odd numbers can be 4k+1,4k1+1,4k2+1.So sum of them should be in the format of 4(k+k1+k2)+3.So the number should produce a remainder 3 when divided by 4.
So the (a),(c),(e) can be discarded.
And by observation we can also say the square of three consecutive odd numbers should be 3k,3k1+1,3k2+1.
So sum of them will be in the format of 3(k+k1+k2)+2.So the number should give a remainder 2 when divided by 3.
Now in between only (d) satisfies this criterion. 😉
d) its 40.
Yes the answer is B ie 5555...
:wow:
5555 is the answer:)
Kindly help,
How many integers between 1 and 1000, both inclusive, can be expressed as the difference of the squares of two non negative integers ??
Thank you.
is the option b correct???