i am uploading some basic question on NUMBER THEORY..
1)prove that every cube number,when divided by 7,leaves remainder 0,1 or 6
2)prove that every cube number,when divided by 9,leaves remainder 0,1 or 8
3)if N is any prime greater that 3,except 7,show that N^6 leaves remainder 1,when divided by 504
4)show that every number and its cube when divided by 6 leaves the same remainder.
11show that every number and its cube when divided by 6 leaves the same remainder.
This holds only for integers
Let the number be x
x3 - x = x*(x2 -1) = (x-1)*x*(x+1)
Now, take any three consecutive numbers. At least one of them will be even ie divisible by 2. One of the three will be a multiple of 3.
So the product will be divisible by 6.
So x3 - x is divisible by 6 ie x3 and x leave same remainders when divided by 6
211)
if N is divisible by 7,then the remainder is zero...
IF N IS PRIME TO 7 then N^6-1 is a multiple of 7 (fermat)
that means either N^3-1 is a multiple of 7 or N^3+1 is a multiple of 7
so,the remainder is either 6 or 1
212) all numbers(N) are of the form 3k-1,3k or 3k+1
so N^3 = 9n-1,remainder is 8
or N^3=9n,remainder 0
or N^3= 9n+1
i hav doubt whether the method is correct...somebody pls check///
215) show that every even power of any odd number is of the form 8n+1
6) find the unit digit of a) 43^(17) b) 1!+2! +3! +......+1999!