sharank wrote:

If \(3^{10}–n\) is divisible by 4, which of the following could be the value of an integer n?

I. 0

II. 1

III. 5

A. I only

B. II only

C. III only

D. II and III only

E. I, II and III

Posted from my mobile device

Given \(3^{10}–n\) = \(4k\)

I. \(n = 0\), we get \(3^{10}–n\) = \(3^{10}\) = \(3^{4}*3^{4}*3^{2}\)

When \(3^{4}*3^{4}*3^{2}\) is divided by \(4\), we get remainder as \(1\). Hence Not Divisible.

\(n = 0\) is not possible.

II. \(n = 1\), we get \(3^{10}–n\) = \(3^{10}–1\)

we know from above \(3^{10}\) divided by \(4\) leaves a remainder of \(1\)

& \(1\) divided by \(4\) will leave a remainder of \(1\).

Hence \(3^{10}–1\) divided by 4 will leave a remainder \((1 - 1) = 0\). Hence divisible.

\(n = 1\) is possible.

III. \(n = 5\), we get \(3^{10}–n\) = \(3^{10}–5\)

we know from above \(3^{10}\) divided by \(4\) leaves a remainder of \(1\)

& \(5\) divided by \(4\) will leave a remainder of \(1\).

Hence \(3^{10}–5\) divided by 4 will leave a remainder \((1 - 1) = 0\). Hence divisible.

\(n = 5\) is possible.

Answer D.

Thanks,

GyM