show that a number and its cube leaves the same remainder when divided by 6
Asked by | 8th Mar, 2009, 10:29: AM
Suppose a number a is divided by 6 gives a quotient q and remainder r, then we can write,
a=6q+r where r is an integer such that
so r takes the values 0,1,2,3,4 and 5.
we see that cube of each of these numbers leaves a remainder equal to the number itself, when divided by 6.
We see that each of the terms except the last one is a multiple of 6.
In the last term we'll get some quotient say t and a remainder equal to r( as mentioned above)
Hence the answer.
Answered by | 8th Mar, 2009, 12:21: PM
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