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## Comment on

Last Digit of a Large Product## When I apply the method that

Because:

49 to the power of 18, is 9 to the power of 18, has a pattern cycle of 2. Which means that 18/2=9. So the digit number of 49 to the power of 18 has a units digit of the first number in that pattern. Which is 9.

The same applies for 13 to the power of 36. 3 has a pattern cycle of 4. 36/4=8. Which means the units digit of 13 to the power of 36 is equal to the first number in that cycle, which is 3.

So 9x3=27. With a unit digit of 7. Can someone explain why this is different then the above stated question?

## Your calculations are a bit

Your calculations are a bit off.

49^18 has units digit 1, and 13^36 has units digit 1

To see why, let's look for a pattern:

49^1 = 49

49^2 = ??1

49^3 = ??9

49^4 = ??1

We can see that the units digit is 9 when the exponent is ODD, the units digit is 1 when the exponent is EVEN.

13^1 = 13

13^2 = ??9

13^3 = ??7

13^4 = ??1

13^5 = ??3

Here, the units digit is 1 when the exponent is divisible by 4. Since 36 is divisible by 4, we know that 13^36 has units 1.

Here's a similar video: https://www.gmatprepnow.com/module/gmat-powers-and-roots/video/1032

And here's a useful article: https://www.gmatprepnow.com/articles/units-digits-big-powers

## Thanks for the clarity. I was

## Exactly what i did. Now

## I suggest that you read the

I suggest that you read the above article.

If you still have questions, please let me know.

Cheers,

Brent

## I did this a much longer way,

## I think that's a perfectly

I think that's a perfectly acceptable approach. The truth of the matter is that this question type is already time consuming. Your solution might be SLIGHTLY longer, but only by a few seconds.

## @YVONNEGMATPREP2017

you can also use x^0 = 1 to start your cycles

makes it much easier

## Sometimes yes and oftentimes

Sometimes yes and oftentimes no.

If we apply it to EVEN bases, that approach can cause confusion.

For example:

4^0 = 1 (units digit = 1)

4^1 = 4 (units digit = 4)

4^2 = 16 (units digit = 6)

4^3 = 64 (units digit = 4)

4^4 = 256 (units digit = 6)

The pattern is 4-6-4-6-4-6-...., so we ignore the 4^0 value.

If I were you, I wouldn't look at x^0 when determining cycles.

Cheers,

Brent

## I thought I understood the

## Let's examine the units digit

Let's examine the units digit of various powers of 13:

13^1 = 13

13^2 = 169

13^3 = ---7 (we need only focus on the units digit)

13^4 = ---1

13^5 = ---3

13^6 = ---9

13^7 = ---7

13^8 = ---1

etc

Notice that the units digits have a pattern: 3, 9, 7, 1, 3, 9, 7, 1, ...

The cycle = 4

Notice that, if the exponent is a MULTIPLE of 4, then the units digit is 1 (13^4 = ---1, 13^8 = ---1, 13^12 = ---1, etc)

Since 36 is a multiple of 4, we know that 13^36 = ---1

FOLLOW-UP QUESTION: What's the units digit 13^302?

Since 300 is a multiple of 4, we know that 13^300 = ---1

From here, we continue the pattern:

13^301 = ---3

13^302 = ---9

So, the units digit 13^302 is 9

Here's an article that discusses this strategy: https://www.gmatprepnow.com/articles/units-digits-big-powers

## Hey Brent,

Thank you for the video.

I do this problem with a slightly different approaches.

Since 13^36 and 3 has 4 cycles, I divide 36 into 3, which gives me no remainder.

When there is no remainder, does it always equal to 1?

For example, if if it's 13^35, the remainder will be 2, and 3^2= 9, so the digit is 9.

But if remainder is 0, will the unit digit always equal to 1?

## That strategy work to work

That strategy work to work when the base has 3 as its units digit, but it won't work for other powers.

Here's why it works when the base has 3 as its units digit:

13^1 = 13

13^2 = 169

13^3 = ---7 (we need only focus on the units digit)

13^4 = ---1

13^5 = ---3

13^6 = ---9

13^7 = ---7

13^8 = ---1

etc

Notice that, when the exponent is divisible by 4, the units digit is 1.

So, for subsequent powers, we simply multiply 1 by successive 3's.

However, for most other bases, the units digit is NOT 1 when the exponent is divisible by 4.

Cheers,

Brent

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