Is it 6^x and 4^(2x) ?

Here is a question to find the pattern.

no matter what 6 has exponent a positive integers the number will end up with 6

6^1 =6
6^2=36 and so forth

4^2*1=16
4^2*2=256 and so forth

C is the best
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Quantity A: The units digit of 6^x
Quantity B: The units digit of 4^(2x)

Notice that (4^2)^x = 4^(2x), so let's replace Quantity B with its equivalent to get:
Quantity A: The units digit of 6^x
Quantity B: The units digit of (4^2)^x

Evaluate 4^2 to get:
Quantity A: The units digit of 6^x
Quantity B: The units digit of 16^x

Since all positive powers of 6 and 16 will have units digit 6, we can conclude that the two quantities will always be equal.
Answer:

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Please explain what do you think?

A units digit of 6 raised to any positive integer is 6.

6^1 = 6
6^2 = 36

16^2 = 256 etc..

C is the answer

Answer: C
By units digits it means the last right digit,
We assume x is 1,2,3,….., n as it is said that it’s a positive integer

x = 1 A = 6^1 = 6 B=4^2*1=(16)
x = 2 A = 6^2 = 36 B=4^2*2= (16)^2
x = 1 A = 6^3 = 216 B=4^2*3= (16)^3
.
.
.
x = n A = 6^n B=4^2*n= (16)^n

So they both end in 6. And the answer is C.

Because both 6 and 16 end in 6, when they have any positive integer power, they will end with 6, definitely this rule is not true for all other numbers. For instance 2 in different powers might end in 2,4,8.
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i didnt understand

Whenever 4 is raised to an positive even power the units digit will always be 6. So the two quantities are equal. Watch out for zero. It is also an even integer but not positive.

Whatever the exponent was for 6 its unit digits is always going to be 6.

test out yourself: -

6^1 = 6

6^2 = 36

6^3 = 36*6 = ...6
etc...

As for 4 it is going to alternate between 4 and 6.

4^1 = 4

4^2 = 16

4^3 = ..4

4^4 = ..6

4^5 = ..4
etc..

notice for 4 when the exponent is odd it ends with 4, when its even it ends with 6.

Quantity A is always going to end in 6. Quantity B is 4^[(2)(x)] --> even exponent all the time because of the 2 multiplied by the x (odd x even = even)

Therefore Quantity B is always going to end in 6 too.

Quantity A : The units digit of \(6^x\)

To find the units digit of power of 6 we need to check the cyclicity in the units digit of powers of 6

\(6^1\) units’ digit is 6 [ 6 ]
\(6^2\) units’ digit is 3 [ 36 ]
\(6^3\) units’ digit is 3 [ 216 ]

=> Units digit of any positive integer power of 6 is 6
=> The units digit of \(6^x\) = 6 (as x is a positive integer)

Quantity B : The units digit of \(4^{2x}\)

To find the units digit of power of 4 we need to check the cyclicity in the units digit of powers of 4

\(4^1\) units’ digit is 4 [ 4 ]
\(4^2\) units’ digit is 6 [ 16 ]
\(4^3\) units’ digit is 4 [ 64 ]
\(4^4\) units’ digit is 6 [ 256 ]

=> Units digit of any positive odd integer power of 4 is 4
=> Units digit of any positive even integer power of 4 is 6
=> The units digit of \(4^2x\) = 6 (as 2x is a positive even integer)

Clearly, Quantity A(6) = Quantity B(6)

So, Answer will be C!
Hope it helps!

Watch the following video (from 2:48 mins) to learn how to find cyclicity of 3 and other numbers

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Why can't I raise it to the power of 0 (it is a positive integer). Am I missing something here?

ZERO:
1. 0 is an integer.
2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.
3. 0 is neither positive nor negative integer (the only one of this kind).
4. 0 is divisible by EVERY integer except 0 itself.

I suggest you brush your fundamentals Integers properties

https://gre.myprepclub.com/forum/gre-qu ... tml#p51913
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