# Question: Last Digit of a Large Product

## Comment on Last Digit of a Large Product

### When I apply the method that

When I apply the method that was used earlier to calculate the units digit of a big number. I come to the answer of 7.
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

Thanks for the clarity. I was stuck when we had to select the number after calculating the cycles.

### Exactly what i did. Now

Exactly what i did. Now confused

### 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 did this a much longer way, but still got A. I found the cycles of 9, to get the pattern 9,1, 9, 1..etc. It has 2 cycles. Then I found the cycles of 3 to get the pattern 3, 9, 7, 1, 3, 9 7, 1...etc. It has 4 cycles. Then I did 18/2 = 9, so the units digit of 49^18 would have to be 1. Then I did 36/4 = 9, so the units digit of 13^36 would have to be 1. And finally, 1X1 = 1. Is this approach too long for the GMAT?

### 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

@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

I thought I understood the cycles in previous questions so- Using the 13^36 as an eg. Once you find that the cycle is 4, you divide the exponent by 4 which is equal to 9, i am unclear to how the logic unfolds from this point

### 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,

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