What is the units digit of the non-negative integer y?

E - as y can have unit digit 0 ,1 and 6 i m not considering 5 bz if unit digit would be 5 it cannot be divisible by 8 but 0,1 and 6 are the number which give same unit digit on any power- as it has two ans its E

Its E.

Each Statement alone is cleary insufficient.
Combine two statements still insuff because the unit digit can be 0 or 6.

1. y is a multiple of 8.
Here is the unit digits of a possible multiple of 8

{8, 6, 4 , 0 } . Clearly insufficient

2. The units’ digit of y^2 is the same as the units’ digit of y^3.
Clearly insufficient

Combining 1 and 2,

we can say the number could an multiple of 8 having the unit digits of either 0 or 6. Hence insufficient

E) should be the answer

Answer is E.

Statement 1: Unit digit can be 2,4,6,8,0: Insufficient
Statement 2: Unit digit can be 0,5,6 : Insufficient
Statement 1&2: Unit digit can be 0,6 : Insufficient

y is a multiple of 8 - patter - 0, 8, 6, 4, 2, 0, etc.....NOT sufficient
y^2 & y^3 have same units digits - y = 0, 5, 6......Not sufficient

1+2 - common 0 & 6 - Now...it may seem dual answers here. But it is stated that y > 0 and it doesn't say if y has 1 digit or 2 or 3....Y>0 is the only condition.

If we assume y is single digit number...then 0 doesn't fit. 6 is the answer.
If we assume y is double digit number....then 0 & 6 fit the answer.
If we assume y is triple digit number....again 0 & 6 fit the answer.

I think I'll go with E?
OA?

Another vote for E!
1) Not sufficient data.
2) This narrows down to the units digit of y to 5, 6 or 0 since only that means y^2 and y^3 have same units digit. So not sufficient.

Combining the two, the units digit may still be 6 or 0 (16 and 40 both satisfy both conditions). So cannot determine unit digit of y.
Answer = E.

The correct answer is E.

(1) Statment 1
- y=8a (a is a non negative int).
- case 1: y=24
- case 2 y=32
-> not suff.

(2) statment 2
- case 1: 5^2=25 and 5^3=125
- case 2: 10^2=100 and 10^3=1000
-> not suff.

(1)+(2) | Stm. 1 + Stm 2
- notice the followings for y=8*a
a=0 y=0= 0*2^3
a=1 y=8= 1*2^3
a=2 y=16= 2*2^3
a=3 y=24= 3*2^3
a=4 y=32= 4*2^3
a=5 y=40= 5*2^3
a=6 y=48= 6*2^3
a=7 y=56= 7*2^3
a=8 y=64= 8*2^3
a=9 y=72= 9*2^3
a=10 y=80= 10*2^3
- so from the pattern of: a*2^3, it is cleaer we can have many cases.
- for example, look at the values of y for a=0,5,10...
- since their last digit is 0, it is clear that statment 2 hold, and we have endless values for y that holds true for 1+2.
-> not suff.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.