# Is there a power of $2$ whose last digit is a $0$?

I'm looking for a power of $$2$$ that ends in $$0$$, so $$2^n = k\cdot 10$$, where $$n \in \mathbb{Z}$$ and $$k \in \mathbb{Z}$$. It doesn't seem to exist. As an equation with two variables, I was looking for a second equation, but just can't think of any.

I also trialled a few consecutive powers of $$2$$ and noticed that there seems to be a sequence perpetually ending in $$2, 4, 8, 6, 2, 4, 8, 6, \cdots$$

How can I prove mathematically that there will never be a power of $$2$$ that ends in $$0$$? Is there a simply way to explain that?

• $5$ divides the RHS, but not the LHS. Commented Sep 24, 2021 at 15:04
• You already proved that the ending digit cycles and $0$ does not appear. Nothing more has to be done. Of course, you can also apply the above comments. Commented Sep 24, 2021 at 15:23
• @Peter I haven't proven it. All I wrote was, that I observed ("noticed") a sequence. Commented Sep 25, 2021 at 14:47
• Thanks for all answers (permanently or temporarily submitted). My first post to this site seemed to have sparked some interest. Commented Sep 25, 2021 at 15:03
• Commented Oct 15, 2021 at 16:03

The only numbers, in the decimal basis, with a digit of $$0$$ for the units, are those that are multiples of $$10$$, a fortiori, multiples of $$2$$ AND $$5$$. Through the fundamental theorem of arithmetic, you can show that there is no power of $$2$$ that is divisible by $$5$$; thus, there is no power of $$2$$ which ends by a $$0$$ digit (in the decimal basis).

• I've just realised that you already cite the fundamental theorem of arithmetic. (+1) I've deleted my answer. Commented Sep 24, 2021 at 19:07
• Thanks. That's the simplest of answers. Though not quite a proof I accept this as the best answer. Commented Sep 25, 2021 at 15:02

Proof 1: Let $$a=2^n$$ and $$b = 2^{n-1}$$. Since $$a = 2b$$, if the last digit of $$a$$ is $$0$$, then the last digit of $$b$$ must be $$0$$ or $$5$$. Since $$b$$ is odd, the last digit can't be $$5$$. But if the last digit of $$b$$ is $$0$$, we can apply the same logic to find that the last digit of $$2^{n-2}$$ is $$0$$, and so on all the way down to $$2$$, but obviously the last digit of $$2$$ isn't $$0$$.

Proof 2: Suppose we take a power of two and write it as $$10a+b$$, and then multiply it by $$16$$. $$16*(10a+b)=160a+16b=160a+10b+5b+b$$. Now clearly the $$160a$$ and $$10b$$ terms don't affect the last digit, and since $$b$$ is even, neither does $$5b$$. So the last digit is $$b$$. This shows that whenever you take a power of two, and then take four more powers of two (that is, multiply by $$2^4=16$$), you don't change the last digit.

Proof 3: The last digit of a number is the same as the remainder when divided by $$10$$, and taking the remainder when divided by some number is known as "modular arithmetic". In modular arithmetic, the set of all numbers with the same remainder is an equivalence class, and these equivalence classes act like numbers. We can apply addition and multiplication to them and get an equivalence class as an answer. For instance, if we multiply a number with a last digit of $$2$$ with a number with a last digit of $$3$$, the product will have a last digit of $$6$$. If we represent the set of numbers with last digit $$r$$ as $$\bar r$$, we can say $$\bar 2 \times \bar 3 = \bar 6$$, and this is true regardless of which numbers we pick from the $$\bar 2$$ and $$\bar 3$$ equivalence classes. So you're asking for an $$n$$ such that the $$\bar 2^n =\bar 0$$. It's easy to see by inspection that this is not possible, but there's also a theorem that if we define $$\phi(n)$$ as the number of numbers less than $$n$$ that are coprime to $$n$$, then $$\bar a^{\phi(n)+1}$$ in $$n$$-modular arithmetic is $$\bar a$$. Since $$\phi(10)=4$$ ($$1,3,7,9$$ are coprime to $$10$$), $$\bar 2^{n+4}=\bar2^n$$.

• I really liked the Proof 2! $(+1)$ Commented Sep 26, 2021 at 22:52

The below tells us more generally what the last digit can be (although the proof is a bit informal).

(The lemma, and a bit more formal reasoning, is really the only thing you were missing.)

Lemma: when multiplying 2 positive integers, only the last digits of each affects what the last digit of the result will be.

Proof: ($$x_i$$ represents the $$i$$-th digit of the integer $$x$$ and $$x_1x_2...x_{n-1} = 0$$ if the number only has 1 digit)

$$a_1a_2...a_n * b_1b_2...b_n$$
$$=(10 * a_1a_2...a_{n-1} + a_n) * b_1b_2...b_n\hspace{3cm}(1)$$
$$=a_n * b_1b_2...b_n + 10 * a_1a_2...a_{n-1} * b_1b_2...b_n$$
$$=a_n * b_n + 10 * a_n * b_1b_2...b_{n-1} + 10 * a_1a_2...a_{n-1} * b_1b_2...b_n\hspace{1cm}\text{(similar to 1)}$$
$$=a_n * b_n + 10 * (a_n * b_1b_2...b_{n-1} + a_1a_2...a_{n-1} * b_1b_2...b_n)$$

The last digit of 10 times any positive integer will always be 0. If you sum such an integer with any other positive integer, the last digit of the result will be equal to the last digit of the other integer. Thus $$a_n * b_n$$ (that is, the last two digits of each integer) is the only thing that affects the last digit of the result. Q.E.D.

Now let's consider the first few powers of 2:

$$2^0 = 1$$ $$2^1 = 2$$ $$2^2 = 2 * 2 = 4$$ $$2^3 = 4 * 2 = 8$$ $$2^4 = 8 * 2 = 16$$ $$2^5 = 16 * 2 = 32$$

Since only the last digit of the current power affects the last digit of the next power, we now have a cycle.

We know any number ending in 2 multiplied by 2 will end in 4. Multiply that by 2 and it will end in 8, then 6, then 2 again. And it will repeat like this indefinitely.

Therefore, the only last digits possible, from $$2^1$$ onward, are 2, 4, 6 and 8.

Negative powers of 2 trivially cannot have a last digit of 0.

Therefore, there is no integral power of 2 having a last digit of 0.

As a bonus result, this approach also shows that the last digit of $$2^{1+i}$$ equals $$2^{1+(i\bmod 4)}\bmod 10$$ for any $$i \ge 0$$.

I'm a Computer Science teacher and was also intrigued by this, so wrote the following program to prove that there are no numbers that are multiples of 5 and powers of 2. I'm definitely no mathematician, so forgive the brute force approach. As stated above by posters, the repeating end digit sequence of 2,4,8,6 is the proof / 'give away'.

# part 1 - set up powers of 2 for reference

maxNumber = int(input('Enter highest number ')) result = 0 power = 0 base2 = [] while True: result = 2**power if result <= maxNumber: base2.append(result) power += 1 else: break print (base2)

# part 2 - go search

found = False for i in range(1,maxNumber+1): if i % 5 == 0 and i in base2: print (i,'is a multiple of 5 and power of 2') found = True if not found: print ('There are no numbers between 1 and',maxNumber,'that are a multiple of 5 and power of 2')