Try This Number Theory Problem

Number Theory Level 1

What is the last digit of $2^9?$

2 4 6 8

5 solutions

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Calvin Lin Staff
Jun 26, 2014

The last digit is 2.

If you observe the pattern of red digits, you see that it repeats in the form 2, 4, 8, 6. Hence, the next value is 2. (How would you prove that this pattern exists?)

Also, if you know that $2^9 = 512$ , then the last digit is 2.

One prove is the 2's multiplication. First you have 2.

$2\times 2=4$

$2\times 2 \times 2=8$

$2\times 2 \times 2\times 2=16$

In the 16 the 6 when multiplied by 2 results in 12. You have 2 again. This 2 makes a 4, the 4 makes a 8 and the 8 makes a 16. The 16 makes 32, because the 6 becomes 12.

Victor Paes Plinio - 6 years, 11 months ago

Ash Bappan
Mar 27, 2015

Simple . If you multiply 256 with 2 , obvious its going to be 2 at the end.

Moderator note:

That's one way of seeing it. But how would you find the last digit of something far much larger, say $2^{123456}$ ?

Srinagini Sravista
Oct 14, 2014

we can observe the last digit it is repeating as 2,4,8,6

Moderator note:

Your solution is incomplete. You did not explain why the digits repeats? Why would the last digits of $2^{1001}, 2^{1002}, 2^{1003}, 2^{1004}$ be $2,4,8,6$ as well instead of something else?

Godwin Bentum
Mar 28, 2015

it 2 multiplying itself 9 times which gives us 512 hence 2 the last digit of the product value

That's one approach. Do you know what to do if the power is large and hard to calculate? For example, what is the last digit of $2 ^ { 999 }$ ?

Calvin Lin Staff - 6 years, 2 months ago

Nathan Blanco
Jul 1, 2014

The pattern is highlighted in red. The pattern is 2 4 8 6. And if you still do not trust that, then just use a calculator. But either way, the answer is 2.

Moderator note:

We often try to avoid using tables or calculators if one isn't needed.

That's perfect

Mostakim Joy - 6 years, 11 months ago

Try This Number Theory Problem

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