x, y, and z are consecutive positive integers such that x

Question

x, y, and z are consecutive positive integers such that x < y < z. If the units digit of x^2 is 6 and the units digit of y^2 is 9, what is the units digit of z^2?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 5

BrentGMATPrepNow · Answer

Let's find some values that satisfy the given information.

If the units digit of x^2 is 6, then it could be the case that x = 6 (since 6^2 = 3 6 )
Since x is the first consecutive integer, it could be the case that y (the second integer) equals 7. 
Notice that y = 7 satisfies the condition that the units digit of y^2 is 9 (since 7^2 = 4 9 )

Since x = 6 and y = 7 satisfy the given information, we now know that z COULD equal 8, in which case z^2 = 8^2 = 6 4 
So, the unit's digit of z^2 must be  4

Answer: D

Hunter2021 · Answer

Is there a more scientific method for solving or proving this? If I try the values from 1-9 it quickly gets clear what we are after but what if the values would go far beyond 10 how would I solve it then?
Thanks a lot for a reply!

ManifestDreamMBA · Answer

Since we are working with only the units digit, the approach will work fine for larger values as well.

deva1 · Answer

Let's check the unit digit patterns:
4^2= 16-> ends in  6 
6^2= 36-> ends in  6 
So, the num's ending in 4 or 6 give squares ending in 6. So, the possible values of x have unit digits: 4 or 6.

Now for y^2 ending in unit digit 9.
3^2= 9
7^2 49-> ends in  9 
So, the possible unit digits for y is 3 or 7.

Now, let's test it for consecutive integers: x,y,z.

Let's test x=14
x^2= 196-> ends in 6,
Then y=15, y^2= 225-> ends in 5 not valid.

So, let's try x=16, y=17, and z=18.
16^2= 256-> ends in  6 
17^2= 289-> ends in  9 
18^2= 324-> ends in  4

It works
So, answer is D.

Hope it helps

x, y, and z are consecutive positive integers such that x < y < z. If

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