Unit digit and power

If $a$ is a positive integer, prove that

$(2^{10})^a \bmod{100} = \begin{cases} 76, a \text{ even} \\ 24, a \text{ odd} \\ \end{cases}$

#NumberTheory

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

This techniques seem interesting. Could you illustrate this with an example?

Agnishom Chattopadhyay - 4 years, 10 months ago

For example : 2^199 Solution: 16^119 =(2^4)^119 =2^796 =2^(10×79+6) =(2^10)^79+2^6 =24 × 64 (here a is odd,by applying the above rule) =4×4 =16 Hence 16 is the last digit place. Check it out by using calc.

MaFiA maNiAc - 4 years, 10 months ago

The last two digits of $2^{199}$ is $88$ , and the last two digits of $16^{119}$ is $36$ .

Efren Medallo - 4 years, 10 months ago

Sorry question was mistyped by me the question is 16^199. Sry guys

MaFiA maNiAc - 4 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$