z is a positive integer and multiple of 2; p = 4z, what is the remaind

Question

z is a positive integer and multiple of 2; p = 4^z, what is the remainder when p is divided by 10?

A) 10
B) 6
C) 4
D) 0
E) It Cannot Be Determined

Source: Platinum GMAT 
 Kudos for a correct  solution .

GMATinsight · Answer

CONCEPT : Remainder when any Number is divided by 10 is same as the Unit digit of the Number

Question : What is the remainder when p = 4^z is divided by 10?  is same as

Question : What is the Unit Digit of p = 4^z?

Given: z is an even Integer

The cyclicity of Unit Digit of 4 is 2 i.e. Unit digit of powers of 4 repeat after every two powers in the pattern {4, 6, 4, 6, 4, ....} 
 4^1 = 4 
 4^2 = 16 
 4^3 = 64 
 4^4 = 256

i.e. Every Even Power of 4 gives the Unit digit = 6 = Remainder

Answer: Option  B

KS15 · Answer

Let z=2 or 4 or 6 etc
Then, p=4^2 or 4^4 or 4^6
Now, 4^2=16 and this divided by 10 gives a remainder of 6
4^4=256 and this divided by 10 also gives a remainder of 6
Answer B

reto · Answer

If p = 4^2, the remainder of 16/10 is 6
If p = 4^4, the remainder of 256/10 is 6

Answer B.

alefal · Answer

Since every number elevated at the at a series of exponents shows a pattern, in this case we spot it from the first three number tested (numbers allowed by the constrains):

- 4^2=16
- 4^3=256
- 4^4=4096

The reminder is always 6.

balamoon · Answer

z is a positive integer and multiple of 2; p = 4^z, what is the remainder when p is divided by 10?

A) 10
B) 6
C) 4
D) 0
E) It Cannot Be Determined

Solution -

z is a positive integer and multiple of 2 -> z=2n

Substitute z in the equation -> p=4^z = 4^2n = 16^n

For any any value of positive integer n, unit digit of p = 6 . Hence the remainder when divided by 10.

ANS. B

Thanks

vishwaprakash · Answer

Z is positive and multiple of 2
Assume Z = 2K

P = 4^Z =  ( (2^2) ) ^ 2K  =  2^(4K)

As per the cyclicity of 2
2^1 -> unit digit 2
2^2 -> unit digit 4
2^3 -> unit digit 8
2^4 -> unit digit 6

Hence P will have unit digit as 6, Hence when divided by 10 will leave remainder as 6

Option B

Turkish · Answer

I will pick two numbers 2 and 4. When I do I get 6 . Answer B

lipsi18 · Answer

Z =4,6,8,10……..
P= 4^z ; 
4^1 =4
4^2=16
4^3 =64
4^4 = ..6 (unit digit will be 6)
Likewise the pattern shows that even power have unit digit as 6 and odd power has unit digit as 4.
Hence 4^(even power) = unit digit =6; and 6/10 =reminder =6

Hence answer is B
Thanks,

LaxAvenger · Answer

p = 2^2z .. cycle of 2: 2, 4, 8 , 6

2^2z will always have 6 as the last digit as it will always have some power of 2^2 in it.. Answer B

Bunuel · Answer

Platinum GMAT Official Solution:

It is essential to recognize that the remainder when an integer is divided by 10 is simply the units digit of that integer. To help see this, consider the following examples:
4/10 is 0 with a remainder of 4
14/10 is 1 with a remainder of 4
5/10 is 0 with a remainder of 5
105/10 is 10 with a remainder of 5

It is also essential to remember that the z is a positive integer and multiple of 2. Any integer that is a multiple of 2 is an even number. So, z must be a positive even integer.

With these two observations, the question can be simplified to: "what is the units digit of 4 raised to an even positive integer?"

The units digit of 4 raised to an integer follows a specific repeating pattern:
4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256
4^(odd number) --> units digit of 4
4^(even number) --> units digit of 6

There is a clear pattern regarding the units digit. 4 raised to any odd integer has a units digit of 4 while 4 raised to any even integer has a units digit of 6.

Since z must be an even integer, the units digit of p=4^z will always be 6. Consequently, the remainder when p=4^z is divided by 10 will always be 6.

In case this is too theoretical, consider the following examples:
z=2 --> p=4^z=16 --> p/10 = 1 with a remainder of 6
z=4 --> p=4^z=256 --> p/10 = 25 with a remainder of 6
z=6 --> p=4^z=4096 --> p/10 = 409 with a remainder of 6
z=8 --> p=4^z=65536 --> p/10 = 6553 with a remainder of 6

Answer: B.

NaeemHasan · Answer

Can´t we simplify 4^z to 2^2z and then solve accordingly? I know that not simplifying is better but it just arised in my mind and solving accordingly i:e; by simplifying I did not get the answer.
Want explanation from experts like Bunuel.

Bunuel · Answer

Yes,  4^z=2^{2z} . Since z is even we can further write this as  2^{2*2k}=2^{4k} .

NaeemHasan · Answer

Thank you for your awesome explanation.

GMAT0010 · Answer

Let's solve this in an "easy peasy" way!
First z = 2k (even integer >0)
then p = 2^2z ==> 2^(4k) which is 16^k. Now : notice that for any number that needs to be divided by 10, the remainder obtained is the unit digit. Here for any value of k, the unit digit will be 6. ( 16, 256, (256*16 = ****6) etc)

So remainder is 6.

Hope this helps!

z is a positive integer and multiple of 2; p = 4z, what is the remaind

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z is a positive integer and multiple of 2; p = 4z, what is the remaind

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