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# If x, y and m are positive integers, and

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Joined: 31 Jul 2017
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If x, y and m are positive integers, and  [#permalink]

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19 Sep 2018, 01:47
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Difficulty:

15% (low)

Question Stats:

80% (01:01) correct 20% (01:56) wrong based on 41 sessions

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$$n = (xy)^m$$. If x, y and m are positive integers, and $$x^2+ y^2= 29$$, then what is the unit's digit of n?

1. 4
2. 3
3. 2
4. 1
5. 0

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Re: If x, y and m are positive integers, and  [#permalink]

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19 Sep 2018, 01:55
rahul16singh28 wrote:
$$n = (xy)^m$$. If x, y and m are positive integers, and $$x^2+ y^2= 29$$, then what is the unit's digit of n?

1. 4
2. 3
3. 2
4. 1
5. 0

First thing that comes to mind is that there are too many variables and we need too specific an information - the units digit of n.
But this is a PS question so if I can find one set of values that satisfies the given constraints, I should get the answer.

$$x^2+ y^2= 29 = 25 + 4$$
So x = 5 and y = 2 satisfy.

$$n = (5*2)^m$$
Say m = 1, then n = 10
Units digit is 0.

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Re: If x, y and m are positive integers, and  [#permalink]

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19 Sep 2018, 11:24
1
Top Contributor
rahul16singh28 wrote:
$$n = (xy)^m$$. If x, y and m are positive integers, and $$x^2+ y^2= 29$$, then what is the unit's digit of n?

A. 4
B. 3
C. 2
D. 1
E. 0

GIVEN: x² + y² = 29

Let's examine some relevant squares of integers:
1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
This as far as we need to go, since all of the other squares are greater than 29

So, which pair of squares will add to 29?
Well, 25 + 4 = 29
In other words, 5² + 2² = 29
So, EITHER x = 5 and y = 2 OR x = 2 and y = 5
It doesn't matter, because we need to find the PRODUCT xy and in both cases, xy = 10

So, the question, "What is the units digit of n = (xy)^m?" becomes "What is the units digit of n = 10^m?"
Since m is a POSITIVE INTEGER, we know that 10^m will always have 0 as its units digit

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Re: If x, y and m are positive integers, and &nbs [#permalink] 19 Sep 2018, 11:24
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