4^1, 4^2, 4^3..

if the number is of the form x^y, where x=4 and y divisible by 4 (y could be 0, 4, 8, 12......), then official answer doesnt make sense, 1, 2, 3 are not divisible by 4.

i would do it like you did. am confused too. what book was that?

semidevil,

are we missing something...could there be a typo? does the question say " y is not divisible by sq of even prime, ie 4" ?then the values of y taken as 1, 2, 3.. might work.

just a thought..

scoobydooby wrote:4^1, 4^2, 4^3..

if the number is of the form x^y, where x=4 and y divisible by 4 (y could be 0, 4, 8, 12......), then official answer doesnt make sense, 1, 2, 3 are not divisible by 4.

i would do it like you did. am confused too. what book was that?

probably helps if I post more info on the question.

the question asks out of 5 choices, which one is a possible units digit of x^y.
the official explanation explains that if you do this:

4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256,
etc
etc,

then looking at the pattern, the units digit will have to be either 4 or 6. I dont remember the answer choices but I think 6 was an answer choice. 4 was not, so 6 was the correct answer.

My question is why did they establish the pattern by testing
4^1, 4^2, 4^3, 4^4, etc etc, and not use 4^8, 4^16, 4^24?

I had to use a calculator to determine that the outcome is the same, but If I saw this on test day, I would have probably tried to manual calculate 4x4x4x4x4x4x4x4x4 etc etc.....

4^4, 4^8, 4^16, 4^24 .. all will have units digit of 6.

This is how I did -

we know that
4*4 = 16, and also
16*16 = (25)6(ie 6 in units digit) since 6*6=36 ..
so 4^4 = 6 in units digit.

Similarly, 4^8 = 4^4 * 4^4
= 6 in units digit * 6 in units digit
= 6 again in units digit.

I hope the above explanation is clear.

Anyone has a better explanation/way ?

oh...they must have used 4^1, 4^2, 4^3, 4^4 to explain the concept that the last digits repeat in a cycle, ie the last digits are 4 and 6 alternately starting from 4^1 and that 4^2; 4^4 have the same last digits.

we can use this cyclic property to establish the last digits, without manually multiplying them out.

4^2=16.

4^4 =(4^2)^2 gives last digit 6. (6 multiplied by itself leaves the last digit as 6)

4^8= (4^2)^4 ; 4^12=(4^2)^6; 4^16 etc etc.....all give last digit 6

From the question, we will know that:
x=4
y=4n (n is positive integer, because y/4=n)

Therefore, x^y=4^(4n)=256^n

It's easy to see that unit digit of 256^n is 6.

the question asks: "If y is divisible by the square of an even prime #..." So this would be 2^2(=4)/y.
Why can't y be 2 (4/2=2)? 2 is divisible by 4 as well.

a is divisible by b means that b divides evenly into a.
In math terms:
a/b = integer.
In other words, a is a MULTIPLE OF b.

The Lost Spaniard wrote:the question asks: "If y is divisible by the square of an even prime #...

y is divisible by the square of an even prime number means that y is divisible by 2² = 4.
In other words, y is a MULTIPLE OF 4:
4, 8, 12, 16, 20...

" So this would be 2^2(=4)/y.

The portion in red implies that 4 is divisible by y -- the OPPOSITE of what we want.
y is divisible by 4 implies the following:
y/4 = integer.

There is only 1 even prime no i.e. 2.
So y = 4k and x = 4
Then xy = 44k will have a unit digit of 6 because even powers of 4 always have unit digit 6.