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If n is a two-digit number, in which n = x^y. If x + y < 8,

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If n is a two-digit number, in which n = x^y. If x + y < 8, [#permalink] New post 26 Oct 2017, 23:49
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If n is a two-digit number, in which n = x^y. If x + y < 8, and x and y are positive integers greater than one, then the units digit of n could be which of the following?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

(F) 5

(G) 6

(H) 7

(I) 8

(J) 9

Kudos for correct solution.


[Reveal] Spoiler: OA
(B), (C), (E), (F), (G) and (H).
[Reveal] Spoiler: OA
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Re: If n is a two-digit number, in which n = x^y. If x + y < 8, [#permalink] New post 27 Oct 2017, 05:00
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x and y are strictly greater than 1 and their sum is less than 8, thus we are left with numbers 2, 3, 4, 5 so as they are all above 1 and their sum is 7 at maximum. Then, we can mix them using the fact that n = x^y.
In this way,

2^4 = 16 so (G)
2^5 = 32 so (C)
3^3 = 27 so (H)
3^4 = 81 so (B)
4^3 = 64 so (E)
5^2 = 25 so (F)

since we have to look at the unit digits!
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Re: If n is a two-digit number, in which n = x^y. If x + y < 8, [#permalink] New post 27 Oct 2017, 10:52
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Bunuel wrote:
If n is a two-digit number, in which n = x^y. If x + y < 8, and x and y are positive integers greater than one, then the units digit of n could be which of the following?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

(F) 5

(G) 6

(H) 7

(I) 8

(J) 9

Kudos for correct solution.


[Reveal] Spoiler: OA
(B), (C), (E), (F), (G) and (H).


Here

X cannot be greater than 5 , because if X= 6 then y can neither be 1 nor 2 as this will nullify the statement.

therefore the possible values are

x=2 then y=5,4 i.e \(2^5\), \(2^4\) = 32 , 16 (\(2^3\), \(2^2\) are not two digit number)

x=3 then y=3,4 i.e \(3^3\) , \(3^4\) = 27, 81

x=4 then y=2,3 i.e \(4^2\) , \(4^3\) = 16 , 64

x=5 then y=2 i.e \(5^2\) = 25

Thus the unit digit are = 1,2,4,6,7
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Re: If n is a two-digit number, in which n = x^y. If x + y < 8,   [#permalink] 27 Oct 2017, 10:52
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If n is a two-digit number, in which n = x^y. If x + y < 8,

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