is it A? stmt 1 alone is sufficient.

since it's asking only the unit digit of n, then just looking at the unit digits:
243^x -->3^x
463^y -->3^y

so from the two lines: 3^x*3^y = 3^(x+y)
if x+y=7, we know thd unit digit of 3^7, so unit digit of n can be found

stmt 2, we know x, but don't know y, so not sufficient

capnx wrote:is it A? stmt 1 alone is sufficient.

since it's asking only the unit digit of n, then just looking at the unit digits:
243^x -->3^x
463^y -->3^y

so from the two lines: 3^x*3^y = 3^(x+y)
if x+y=7, we know thd unit digit of 3^7, so unit digit of n can be found

stmt 2, we know x, but don't know y, so not sufficient

Capnx,

I'm just curious about your solution,...because it looks like you have simplified 243 down to 3, but 463 does not have "3" as a factor. So how did you finish the problem?

whether 3 is a factor is not needed. 3 is the last digit, so to get the last digit of the product of two numbers, all you need is the last digit.

ie: 13^x = ______
x = 1, = 13 --> 3^1=3
x = 2, = 169 --> 3^2=9
x = 3, = 2197 --> 3^3=27

so the power of the last digit only depends on the power of the last digit.

Thanks for the clarification,... this, again, is the premiere problem I am having (as I posted on another forum question).

I am always missing the point of finding out whether the possibilities are "sufficient" versus knowing the actual answer. I need to burn this in my brain.

I would go with A.

Capnx makes good points above with his explanation. Another way to do it would be to plug in values, for the non-algebraic way. We can see that all we need to do is find out how many times 3 is multiplied, so I made a list to find out when 3 repeats. Since the question only cares about the units digits, we can apply the last digit shortcut. 3^1 = 3, 3^2=9, 3^3=27, 3^4=81, 3^5= 243, 3^6=729 so we're sure we're back at the beginning, so we can just repeat as needed. Obviously don't multiply things out, but I did it to just illustrate the pattern.

Statement 1: x + y = 7

If x=1 and y=6, the last digit would be (3^1=3) * (3^6=729) = 3*9= 27 (using the last digit shortcut)

If x=2 and y=6, the last digit would be (3^2=9) * (3^5=243) = 9*3= 27

If x=3 and y=4, the last digit would be (3^3=7) * (3^4=81) = 7*1= 7

and so on....we can see that 7 will be the final digit for all of them. Sufficient.

Statement 2: x = 4 Doesn't tell us anything about y, thus insufficient.

IMHO A

we know, that

3^1=3(9 is a unit digit)
3^2=9(9 is a unit digit)
3^3=27 (7 is a unit digit)
3^4=81 (1 is a unit digit)
3^5=243(3 is a unit digit)

so , now we r moving to stmnt 1

x+y=7

since we know that x and y are positive integers, then x colud be equal to 3;5;4;2 ,then y will be equal to 4;2;3;5 respectively.

so if x =3 ,and y =4 then the units digit of n will be 7 (since 3^3=27 (7 is a unit digit)
3^4=81 (1 is a unit digit) 7*1 =7)

if x=5 then y=2 ,so the units digit of n will be 7 again (since 3^2=9(9 is a unit digit) and 3^5=243(3 is a unit digit) ; 9*3=27)

A is sufficient

B is not sufficient, since we have no info about y.

imho ,the trick of this question is the tendency of choosing C.

by the way, this question is time -consuming ( what is the source of it?

you are listing choices (variations of x+y=7) and then arbitrarily decide to choose x=3 and y=4, perhaps after grueling solution and understanding that all variations give the same answer.

Instead note please, the unit's digit of n will be resolved by finding x and y
st(1) suggests that x+y=7 which is only possible if x or y is at least 1 and/or x or y is at most 6 (x,y are positive integers per this q.). So the variations we have (3^1)*(3^6) or two more variations, like ((3^2)*(3^5), (3^3)*(3^4) listed by you can be quickly regrouped as 3^7, because we account for the last number=3 in both factors 243 and 463 and be it (3^1)*(3^6) or (3^3)*(3^4) this can be rewritten as 3^(x+y=7)

this way you save ton's of time and avoid lengthy solution for all variations, not mentioning arbitrary choice "let x=3 and y=4"

st(2) x=4 and y can be anything is Not Sufficient

rainmaker wrote:Please explain your answer:

If (243)^x(463)^y = n, where x and y are positive integers, what is the units digit of n?

(1) x + y = 7

(2) x = 4

where ^ means "raised to the power of"

LalaB wrote:IMHO A

we know, that

3^1=3(9 is a unit digit)
3^2=9(9 is a unit digit)
3^3=27 (7 is a unit digit)
3^4=81 (1 is a unit digit)
3^5=243(3 is a unit digit)

so , now we r moving to stmnt 1

x+y=7

since we know that x and y are positive integers, then x colud be equal to 3;5;4;2 ,then y will be equal to 4;2;3;5 respectively.

so if x =3 ,and y =4 then the units digit of n will be 7 (since 3^3=27 (7 is a unit digit)
3^4=81 (1 is a unit digit) 7*1 =7)

if x=5 then y=2 ,so the units digit of n will be 7 again (since 3^2=9(9 is a unit digit) and 3^5=243(3 is a unit digit) ; 9*3=27)

A is sufficient

B is not sufficient, since we have no info about y.

imho ,the trick of this question is the tendency of choosing C.

pemdas wrote: this can be rewritten as 3^(x+y=7)

pemdas,I got what u said .that was nice ) sometimes life is easier than my brain can imagine )

HI

Is 463 a prime number??

HSPA wrote:HI

Is 463 a prime number??

Yup.

pemdas,

Your solution is ofcourse the fastest one....thank you

pemdas wrote:you are listing choices (variations of x+y=7) and then arbitrarily decide to choose x=3 and y=4, perhaps after grueling solution and understanding that all variations give the same answer.

Instead note please, the unit's digit of n will be resolved by finding x and y
st(1) suggests that x+y=7 which is only possible if x or y is at least 1 and/or x or y is at most 6 (x,y are positive integers per this q.). So the variations we have (3^1)*(3^6) or two more variations, like ((3^2)*(3^5), (3^3)*(3^4) listed by you can be quickly regrouped as 3^7, because we account for the last number=3 in both factors 243 and 463 and be it (3^1)*(3^6) or (3^3)*(3^4) this can be rewritten as 3^(x+y=7)

this way you save ton's of time and avoid lengthy solution for all variations, not mentioning arbitrary choice "let x=3 and y=4"

st(2) x=4 and y can be anything is Not Sufficient

rainmaker wrote:Please explain your answer:

If (243)^x(463)^y = n, where x and y are positive integers, what is the units digit of n?

(1) x + y = 7

(2) x = 4

where ^ means "raised to the power of"

LalaB wrote:IMHO A

we know, that

3^1=3(9 is a unit digit)
3^2=9(9 is a unit digit)
3^3=27 (7 is a unit digit)
3^4=81 (1 is a unit digit)
3^5=243(3 is a unit digit)

so , now we r moving to stmnt 1

x+y=7

since we know that x and y are positive integers, then x colud be equal to 3;5;4;2 ,then y will be equal to 4;2;3;5 respectively.

so if x =3 ,and y =4 then the units digit of n will be 7 (since 3^3=27 (7 is a unit digit)
3^4=81 (1 is a unit digit) 7*1 =7)

if x=5 then y=2 ,so the units digit of n will be 7 again (since 3^2=9(9 is a unit digit) and 3^5=243(3 is a unit digit) ; 9*3=27)

A is sufficient

B is not sufficient, since we have no info about y.

imho ,the trick of this question is the tendency of choosing C.

D in answer.

1) X+Y =7

Suppose X<Y
x=1, y=6 => units digit = 7
x=2, y=5 => units digit = 7
x=3, y=4 => units digit = 7

As 243 and 463 have the same units digits, the case x>y leads to the same result.

=> stat 1 is sufficient

2) Only know x => insufficient

*Another way: For stat 1): ignore 243 and 463. What matters here, the units digit of n, is exactly the same as the units digits of 243^x*243^y (=243^7) => units digit is 7