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## Comment on

Units Digit of 53 to Power of 35## What do you advise?

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!

Question link: https://gmatclub.com/forum/what-is-the-units-digit-of-the-following-expr...

That's a great approach!

## A simple way of thinking

## Perfect!

Perfect!

## 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

## 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,

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,

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.

## Question link: https:/

Question link: https://gmatclub.com/forum/what-is-the-units-digit-of-the-following-expr...

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!

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