Question

a number which is formed by writing one digit 8 times will be always divisible by

Answer 1

hello friend!

the answer is 2.

it will always be divisible by 2.

thank you....mark as branliest!!

Answer 2

Let N = 44444444......4444444 (2045 times). find remainder when N is divided by 103 ?
44444 mod 103 = 51
any number written p - 1 times div by p where p is prime
so only last 5 digits will be dere so 44444 mod 103=51
Let N = 111...111 ( 73 times ). When N is divided by 259, the remainder is R1, and when N is divided by 32, the remainder is R2. Then R1 + R2 is equal to
A. 6
B. 8
C. 253
D. None of these
Any digit repeated p - 1 times is divisible by p where p is prime
111111 ( repeated 6 times ) is divisble by 259 ( = 37 * 7 ) = > 1111... repeated 72 times is divisible by 259. As there are 73 one's in the number remainder is 1.
OR 111....... 36 times is divisible by 37 = > 1111...72 times is divisible by 37 = > only last 1 will be there giving a remainder of 1 ( we can use either of this method )
For 32 (2^5) we need find the remainder for last 5 digits. General rule for 2^n is to check the last n digits , for remainder.
check 11111 mod 259, we get 7 as the remainder

so sum 7 + 1 = 8, our answer!
What is remainder when 2014 2015 is divided by 121 ?
Use Eulers theorm, 121 = 11 * 11 = > E (121)=121 * 10/11 = 110
now 2014^2015 mod 121 = (-43)^35 mod121
= -43*(43*43)^17 mod 121
= -43*(34)^17 mod 121
= -43*34*(34*34)^8 mod 121
= -43*34*(-54*-54)^4
= -43*34*(12)^4 mod 121
= -43*34*144*23 mod 121 = -43*34*23^2 mod 121 = -43*34*45 Mod 121
= -87 mod 121 = 34.
OR
78 ^ 2015 mod 121 = ( 1+ 77 ) ^ 2015 mod 121= 77C0+2015C1*77 + rest of the terms divisble by 11 = ( 1+ 2015 * 77 ) mod 121 = 34
If m is a natural no. and N = 2m, what is the remainder when N! is divided by 2N ?
highest power of 2 in N! = (2^m)-1 = N -1
N! = 2^(N-1)*k
N! mod 2^N = 2^(N-1) { k mod 2} =2^(N-1)
P is a prime number greater than 30. When P is divided by 30, the remainder is x. How many different values of x are possible ?
a. 9
b. 8
c. 10
d. 11
P = 30n + x, P has to be 6k+1 or 6k-1. Hence x also has to be of the form 6k+1 or 6k-1. Possible values of x are 1,5,7,11,13,17,19,23,25,29. out of these x cannot be 5 or 25. hence 8 values
OR just find number of coprime less than 30 = 30*1/2*2/3*4/5 = 8
If n = 1 + x, where x is the product of 4 consecutive natural numbers, then n is:
I. an odd number.
II. an even number.
III. a prime number.
IV. a perfect square.

1) II only
2) III and IV only
3) I and IV only
4) I only
5) None of these
For 4 consecutive natural number there will be atleast 2 even numbers so product will be even = > x is even and x+1 is odd
so 1 is true and 2nd is false
now to decide where n is prime or not
try for first four natural numbers which give a prduct of 24 and therefor n =25, therefore n is not prime but could be prefect square

OR

n= (a-1)*(a)*(a+1)(a+2)
n= (a^2+a-1)(a^2+a-1)
so n is perfect square

OR