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Senior Manager  P
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 435
x, y, and z are consecutive positive integers such that x < y < z.  [#permalink]

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Difficulty:   15% (low)

Question Stats: 78% (01:07) correct 22% (01:32) wrong based on 37 sessions

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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

I solved this by picking numbers:
1) What can have a units digit of 6 when squared? Try x=6, x^2=36.
2) Looking at y, since they are consec ints, y=7 and 7^2 is 49, which also fits our given information.
3) If x=6,y=7 then z=8, 8^2 = 64 so the units digit is 4, D.

Official explanation:
Because the units digit of x^2 is 6, the units digit of x must be either 4 or 6. Because x, y, and z are consecutive positive integers, the units digit of y is 5 if the units digit of x is 4. However, the units digit of y can’t be 5 because the problem tells us that the units digit of y^2 is 9; if the units digit of y were 5, then the units digit of y^2 would also be 5. Therefore, the units digit of x is 6, the units digit of y is 7, and the units digit of z is 8. The units digit of z^2 , then, is 4.
Director  G
Joined: 09 Mar 2018
Posts: 994
Location: India
Re: x, y, and z are consecutive positive integers such that x < y < z.  [#permalink]

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energetics wrote:
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

Key word: x, y, and z are consecutive positive integers

cases which will satisfy x < y < z will be

6<7<8
16<17<18

$$z^2$$ will end on a 4

D
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Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Senior Manager  S
Joined: 12 Sep 2017
Posts: 302
x, y, and z are consecutive positive integers such that x < y < z.  [#permalink]

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energetics wrote:
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

I solved this by picking numbers:
1) What can have a units digit of 6 when squared? Try x=6, x^2=36.
2) Looking at y, since they are consec ints, y=7 and 7^2 is 49, which also fits our given information.
3) If x=6,y=7 then z=8, 8^2 = 64 so the units digit is 4, D.

Official explanation:
Because the units digit of x^2 is 6, the units digit of x must be either 4 or 6. Because x, y, and z are consecutive positive integers, the units digit of y is 5 if the units digit of x is 4. However, the units digit of y can’t be 5 because the problem tells us that the units digit of y^2 is 9; if the units digit of y were 5, then the units digit of y^2 would also be 5. Therefore, the units digit of x is 6, the units digit of y is 7, and the units digit of z is 8. The units digit of z^2 , then, is 4.

Hi!

Just choose 6, 7 and 8, you have to search for an integer that has a units digit of 6.

Cyclicity rule of 6... it will always finish with 6, so start with that number.

"If the units of If the units digit of x^2 is 6"

Hence...

6*6 = 36
7*7 = 49
8*8 = 64

D x, y, and z are consecutive positive integers such that x < y < z.   [#permalink] 13 Feb 2019, 16:46
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x, y, and z are consecutive positive integers such that x < y < z.

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