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show that a number and its cube leaves the same remainder when divided by 6
Asked by | 08 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.

So,

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)

Answered by | 08 Mar, 2009, 12:21: PM
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