dreambeliever wrote:

What is the value of x if x is the remainder obtained when \(2^{8p+2} + Z\)is divided by 5 and p is a positive integer?

1) Z= 6

2) Z is even

\(2^{8p+2}\)

The unit digit of the above expression will always be 4.

Here's how;

\(2^n\) has a cyclicity of 4. And the unit digits appear as 2,4,8,6 with every increment in n.

When "8p+2" is divided by 4, it always render a remainder of 2.

8p is divisible by 4 leaving the remainder as 0.

When 2 is divided by 4, it will leave a remainder of 2.

Whenever; in expression \(2^n\), n divides by 4 and leaves a remainder of 2, the unit digit of the expression will be 4.

1. Z=6

4+6=0

The expression \(2^{8p+2} + Z\) will always be divisible by 5 as the unit digit will always be 0. x=0;

Sufficient.

2. Z=even;

Z=2; The expression's units digit will be 6 and x=1;

Z=6; The expression's units digit will be 0 and x=0;

Not Sufficient.

Ans: "A"

_________________

~fluke

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