# Question: Units Digit of 53 to Power of 35

## Comment on Units Digit of 53 to Power of 35

Of the (13^5)(15^4)(17^5)
(13)^5 = Odd
(17)^5 = Odd
(15)^4 = Unit digit of 5
5*Odd = Unit of 5

### That's a great approach!

That's a great approach!

### A simple way of thinking

A simple way of thinking about this for me is to find the cycle, in this case 4 (3, 9, 7, 1), then divide the exponent by the cycle, 4. 35/4 = 32 R 3, therefore the answer is the 3rd digit in the cycle. If R is 0 then the answer will be the last digit in your cycle.

Perfect!

### Hi Brent,

Hi Brent,

Very confused with this example. You state that "when N is divisible by 4, the Units Digit of 53^n is 1 given that 53^8 and 53^4 are both divisible by 4 and have a units digit of 1.

However isnt 53^35 divisible by BOTH 5 and 7? We know that 5th units digit in the sequence is "3", whereas the 7th units digit in the sequence is "7".

How would you know to pick 7 over 5 as a divisor of 53^35?

Thank you.

### Let's examine the pattern:

Let's examine the pattern:
53^1 = 53
53^2 = --9
53^3 = --7
53^4 = --1
53^5 = --3
53^6 = --9
53^7 = --7
53^8 = --1
53^9 = --3
53^10 = --9
53^11 = --7
53^12 = --1
53^13 = --3
53^14 = --9
53^15 = --7
53^16 = --1

Notice that the pattern repeats ever 4 powers. So, the pattern has cycle 4.
This is why we focus on the powers that are multiples of 4.
We know that 53 to the powers of ANY multiple of 4 will equal ---1
For example:
53^4 = --1
53^8 = --1
53^12 = --1
53^16 = --1
So, without doing any extra work, I know that 53^400 will have units digit 1.

Conversely, if we focus on powers of 5 (or powers of 7), we see that there is no consistency:
53^5 = --3
53^10 = --9
53^15 = --7
Hmm, so what is the units digit of 53^500? It's hard to tell.

So, once we know the cycle of the pattern, we must focus on multiples of that cycle.

Does that help?

Cheers,
Brent

### This helps a lot. Thank you

This helps a lot. Thank you for the extremely detailed and comprehensible response! :)

### I get it now.

I get it now.

Rather than just stick to the multiplication rule I should identify patterns or behaviors of the cycle.

Thanks Brent.

### Hello Brent,

Hello Brent,

I just used the quotient laws from previous lessons. If we take 53 as X and 35 as "a" and "b" the question simplifies as X^(a.b) that is 53^5*7.

So if 53^5 ends in 3, then any multiple of 5 must end in 3.

### You are correct to say that

You are correct to say that 53^35 = (53^5)^7

However, we cannot conclude that, "since 53^5 ends in 3, then any multiple of 5 must end in 3"

To see why this is not true, let's examine the units digits of several powers of 53:

53^1 = 53
53^2 = --9
53^3 = --7
53^4 = --1
53^5 = --3
53^6 = --9
53^7 = --7
53^8 = --1
53^9 = --3
53^10 = --9
53^11 = --7
53^12 = --1
53^13 = --3
53^14 = --9
53^15 = --7
53^16 = --1

Notice that 53^5 has units digit 3, and 53^10 has units digit 9, and 53^15 has units digit 7.

Does that help?

Cheers,
Brent

### Hey Brent,

Hey Brent,

now one solution seems to be:

Of the (13^5)(15^4)(17^5)
(13)^5 = Odd
(17)^5 = Odd
(15)^4 = Unit digit of 5
5*Odd = Unit of 5

But how is an odd number simply irrelevant?

Cheers.

We know that (13)^5 and (17)^5 are both odd.
So, (13)^5 x (17)^5 = some odd number.
Also, (15^4) has units digit 5

This means (13^5)(17^5)(15^4) = (some odd number)(some number with units digit 5)

If we multiply any odd integer by 5, the units digit is always 5.
For example: (1327)(225) = ------5
And (613)(65435) = ------5
And (807)(1915) = ------5
etc

Does that help?

Cheers,
Brent

Got it, thanks!

### that 500-650 one is not

that 500-650 one is not tricky at all using the strategy.

just has to calculate a bit for the unit digit.

81*5 = 5 81 is 9*9