goelmohit2002 wrote:What is the units' digit of (p^(xy)) ?
1) The units' digit of p^x is 1.
2) The units' digit of p^y is 1.

anyone is sufficient to answer

so D

yes D

p^(xy)=(p^x)^y=(p^y)^x

Let's take an example...

p^x = 81.
y = 0.5

Here (p^(xy))= 9

But when

p^x = 81.
y = 1
(p^(xy))= 81.

So isn't A or B alone insufficient....

Please tell what I am missing here ?

If you guys too agree with my line of reasoning...then kindly tell what should be the correct answer and how to reach there ?

goelmohit2002 wrote:Let's take an example...

p^x = 81.
y = 0.5

Here (p^(xy))= 9

But when

p^x = 81.
y = 1
(p^(xy))= 81.

So isn't A or B alone insufficient....

Please tell what I am missing here ?

If you guys too agree with my line of reasoning...then kindly tell what should be the correct answer and how to reach there ?

u r correct; if x,y both r integer then d, otherwise E, I feel

IMO D
if
p^x = 81
y = 0.5

(p^(xy)) not equal to 9

==> p^(xy) = p^(81/2)

gmatv09 wrote:IMO D
if
p^x = 81
y = 0.5

(p^(xy)) not equal to 9

==> p^(xy) = p^(81/2)

think again

xy = 81 * 1/2 :



what am i missing?

This should be E

p^x x can be 0 or 9 insufficient, no info about y
if x=9 and y=1 2^9=512
if x=0 and y=anything 2^0=1

P^y y can be 0 or 9 insufficient, no info about x
see previous

together x*y is either 81 or 0
pick a value for p (i chose 2)
2^81 is ridiculous but will always be an even number
2^0 is always 1
this means you have two different answers and not enough info to solve the problem

wccotton wrote:This should be E

p^x x can be 0 or 9 insufficient, no info about y
if x=9 and y=1 2^9=512
if x=0 and y=anything 2^0=1

P^y y can be 0 or 9 insufficient, no info about x
see previous

together x*y is either 81 or 0
pick a value for p (i chose 2)
2^81 is ridiculous but will always be an even number
2^0 is always 1
this means you have two different answers and not enough info to solve the problem

Can you please explain it more....basically which steps you are doing using
a) statement 1 alone.....
b) statement 2 alone....
c) combining statement 1 and statement 2.

Currently it looks to be really confusing to figure out the same....

I tried it using the log approach....(although not in course of GMAT....but could not think of any other way)

As we have already discussed statement 1 and statement 2 alone are insufficient....so combining....
p^x = A ( given that p^x has unit digit of 1...so let's take that number as A)
x log p = log A......(3)

p^x = B ( given that p^y has unit digit of 1...so let's take that number as B)
y log p = log B ......(4)

dividing equation 3 and equation 4.....

=> x/y = (log A/log B)
=> x = y(log A/log B)......(5)

putting this value in given question at hand....
p^(x*y) = p^(x^2* (log A/log B)) = (p^(x^2) * (log A/ log B)).......(6)

Now

p^(x^2) has units digit always 1 since p^x has unit digit of 1.

But log A/ log B can have any value..if it is integer then the product in equation 6....will have units digit as 1.....but if log A/ log B is not integer then it can have any value....

so insufficient...even when combined A and B....

Can some one please tell if there is any other better way to solve.....

By putting numbers too it comes out to be E...

Let say x = y = 0.5

Case 1
=====
p = 81 * 81
Here both p^x and p^y has unit digit as 1.
=> p ^ (x * y) = p ^( 0.5 * 0.5)
thus value of unit digit of p is 9......

Case 2
=====
p = 81 * 81 * 81 * 81
Here both p^x and p^y has unit digit as 1.
=> p ^ (x * y) = p ^( 0.5 * 0.5)
thus value of unit digit of p is 1......

Thus two different values...thus insufficient thus....E.

Can someone please tell is there a way to solve this question without using number plugging or the log method as I posted above.

I guess the easiest thing to do is to plug in numbers to see what you can get.

Let's take statement A) The units digit of p^x is 1
So for what values of p and x can the units digit be 1?
If p=3 and x=4, you have 3^4 which is 81 which satisfies this statement. Plug these values back into the equation you get 3^4y which is really 81^y.
So if y = 1 unit get 81^1 which is 81, but if y=0.5, you get 81^0.5 which is equal to 9. This same logic applies to statement B).

You can also see a breakdown if you use negative numbers for x or y, i.e. 81^(-1) has a units digit of 0. (This value is 3^(4*(-1))=3^-4 which has a units digit of 0)

When you take the two statements together, you know p^x units digit is 1 and p^y units digit is 1. This requires a bit more creativity to see if you can create a number with a units digit of 1 that will violate these rules.
If p=1.1, x=2 and y=4 you get the following
A)1.1^2=1.21
B)1.1^4=1.461
and together 1.1^(2*4)=1.1^8 which is 2.14

But if p=3, x=4 and y=4 you get the following
A)3^4=81
B)3^4=81
and together 3^(4*4)=3^16=43,046,721 which has a units digit of 1.

This is a lot easier if there are restrictions on the numbers.

Thanks.

But Can someone please help me to learn the same without relying on number substitution....guessing the numbers in tight test conditions is just too time consuming and prone to errors......

Actually, you should consider that there are some situations where number picking is the only way. This is one situation. There are countless others where you can solve algebraically, but just breaking it down into cases according to x and y is waaay too tedious and not worth it (at least in my head).

Normally, for this type of exercise, you'd use one of the classical rules of units digit:

A number with units digit 1, 5 or 6 will keep its units digit throughout all its >=1 powers.

It's important to remember that rule, so you'll have to check for some <1 powers (as you did with 0.5).

Extra: @wccotton: don't use negative powers when establishing the units digit (unless specifically asked). Most of the time, when we talk about units digits, we only use positive powers. Units digits for non-integers (i.e. 0.5) are usually not discussed.