Playing with 2

what is the last four digits of 2^2016 . provide detailed solution

Calvin: This discussion is now locked as it relates to a recently posed problem.

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`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`	$\sum_{i=1}^3$
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`\boxed{123}`	$\boxed{123}$

Comments

Firstly, $2^{4x} \equiv 1 \pmod{5}$ with every positive integer x (easily proved).

As a result:

$2^{20}-1=(2^4-1)(2^{16}+2^{12}+2^8+2^4+1)$ is divisible by 25, means $2^{20} \equiv 1 \pmod{25}$

$2^{100}-1=(2^{20}-1)(2^{80}+2^{60}+2^{40}+2^{20}+1)$ is divisible by 125, means $2^{100} \equiv 1 \pmod{125}$

Once more and we have: $2^{500} \equiv 1 \pmod{625}$

So: $2^{2000} \equiv 1 \pmod{625}$

$(2^{2000}-1)$ is divisible by 625 -> $(2^{2004}-2^4)$ is divisible by 10000

Then $(2^{2016}-2^{16})$ is divisible by 10000

The last 4 digits of $2^{2016}$ is the same as that of $2^{16}$ , which is 5536.

Ngọc Bùi Tiến - 6 years, 4 months ago

Great. Another way to tell that $2 ^ {500} \equiv 1 \pmod{625}$ is to use Euler's Theorem directly.

Calvin Lin Staff - 6 years, 4 months ago

good solution

Dheeraj Agarwal - 6 years, 4 months ago

Can you post your solution on to this problem? Thanks!

Calvin Lin Staff - 6 years, 4 months ago

What have you tried? Do you know how to find the last digit of a power?

Calvin Lin Staff - 6 years, 4 months ago

i know to find the last digit but i want to know the solution to above problem

Dheeraj Agarwal - 6 years, 4 months ago

Actually there are many ways checking cyclic changes.For more than one last igits mod 10^n acompanied by congruence and Binomial Theorem. Sir, can you give me the most generalized process.

Chandrachur Banerjee - 6 years, 4 months ago

For last digit does check cyclic always work?

Chandrachur Banerjee - 6 years, 4 months ago

Last two digits will be 36

Gaurav Negi - 6 years, 4 months ago

i think its 32

Akhil Pampana - 6 years, 4 months ago

Dhileepan Thillalangadi - 6 years, 4 months ago

0136

Aayush Badade - 6 years, 4 months ago

@Dheeraj Agarwal Please refrain from asking for solutions to problems that were recently posted, especially if the problems already have solutions.

Calvin Lin Staff - 6 years, 4 months ago

Last 2 digits are 36

2^2016 = 2^10^201 × 2^6

= 1024^201 × 2^6

24^odd power × 64

24^even power ends with 76 and 24^odd power ends with 24

so 24 × 64 = 1536 last 2 digits are 36

Sunil Pradhan - 6 years, 4 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}`	$\boxed{123}$