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11 Nov 2021, 03:45
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D
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Difficulty:
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Question Stats:
92% (01:13) correct
9%
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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
(A) 0
(B) 1
(C) 2
(D) 4
(E) 5
11 Nov 2021, 08:07
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Bunuel
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
(A) 0
(B) 1
(C) 2
(D) 4
(E) 5
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 = 36)
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 = 49)
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 = 64
So, the unit's digit of z^2 must be 4
Answer: D
13 Feb 2022, 04:07
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BrentGMATPrepNow
Bunuel
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
(A) 0
(B) 1
(C) 2
(D) 4
(E) 5
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 = 36)
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 = 49)
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 = 64
So, the unit's digit of z^2 must be 4
Answer: D
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!
