What is the units digit of the expression 6^x + 4^(y + z)

Question

What is the units digit of the expression  6^x+4^{y+z} , where x, y and z are positive integers?

(1) y + z is even.
(2) x*y*z = 1

96Chris · Answer

Cyclicity of units digit of 6: 
6, 6, 6, 6.....
Cyclicity of units digit of 4: 
4, 6, 4, 6.....

So for 6^x we will always have a units digit of 6, for 4^y+z a units digit of 4 if y+z is odd, and 6 if y+z is even.

(1) y+z is even. Tells us what we need to know. Units digit will be 6 + 6 = 2
(2) Means x=y=z=1, so y+z = 1+1 = 2 --> Odd power, so units digit is 4. 4 + 6 = 10, so units digit is 0. Sufficient.

IMO D

TestPrepUnlimited · Answer

6^x  always ends in 6, so we want to focus on the other part  4^{y + z} . The last digit cycle for powers of 4's is 4, 6, 4, 6 etc. So we can see it repeats every 2 numbers.

Statement 1:

As we said it repeats, so y + z even means  4^{y + z} = 4^{2*	ext{some integer}} = 16^{	ext{some integer}}  always ends in 6. Then we know the units digit is always the same. Sufficient.

Statement 2:

x, y, and z are all positive integers so the only solution here is x = y = z = 1. Sufficient.

Ans: D

ReedArnoldMPREP · Answer

When you see a question about UNITS DIGITS with EXPONENTS, you should think about the UNITS DIGITS CYCLE.

Every number has a cycle that its exponents move through. Since the units digit is the only digit that determines the units digit of the outcome, you only ever need to worry about it. This problem deals with the units digit cycles of 4 and 6.

6's units digit cycle is very easy. Every exponent of every number that ends in 6... also ends in 6. So 6^x will have a units digit of 6. What we need to know is what the units digit of 4^(y+z) is.

4's units digit cycle looks like this: 4, 6, 4, 6, 4, 6... etc

Every even power ends in a 6, every odd power ends in a 4.

So we'll either add units digits 6+4 (if y+z is odd) or 6 + 6 (if y+z is even). That ends up being all that matters for this question: is y+z even or odd (Note: we're told all variables are POSITIVE INTEGERS).

Statement 1 tells us just what we wondered. It says y+z is odd. So the units digit cycle of '4' will end on '4.'

4+6 will end in a 0.

Statement 2 is also enough. This tells us every variable is 1, so y+z = 2, an even number.

The answer must be D.

TAKEAWAYS:
--When you see exponents and units digit, find the units digit cycle.
--You don't have to (but you could) memorize each units digit cycle for each units digit. It's more important that you be able to figure it out quickly (none of them are longer than a cycle of '4.' For instance the units digit cycle for 3 is: 3, 9, 7, 1.

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5= 243
3^6 = ____9
3^7 = ____7
3^8 = ____1
etc.

--Investigate a question upfront! Often what seems to be asked isn't *really* what's asked.

What is the units digit of the expression 6^x + 4^(y + z)

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What is the units digit of the expression 6^x + 4^(y + z)

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