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# Is √p a non-integer? (1) p = r × 10 r, where r is a positive odd in

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Is √p a non-integer? (1) p = r × 10 r, where r is a positive odd in [#permalink]

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23 Mar 2017, 17:11
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64% (01:01) correct 36% (01:04) wrong based on 105 sessions

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Is $$√p$$ a non-integer?

(1) $$p = r × 10^r$$, where r is a positive odd integer.
(2) $$p = 9 × 10^s$$, where s is a positive integer.
[Reveal] Spoiler: OA

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Director
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Re: Is √p a non-integer? (1) p = r × 10 r, where r is a positive odd in [#permalink]

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23 Mar 2017, 18:10
avinsethi wrote:
A?

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app

Yes its A...

(1) 10^odd integer is never a square .
HEre r is odd integer.

suff.

(2) we dont have enough info of s (even /odd).
if s=even then YES
if s=odd then NO
insuff

Ans A

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VP
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Is √p a non-integer? (1) p = r × 10 r, where r is a positive odd in [#permalink]

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23 Mar 2017, 18:20
Is $$√p$$ a non-integer?

(1) $$p = r × 10^r$$, where r is a positive odd integer.

Let r= 1..........$$√10$$...........Answer is yes

Let r= 3..........$$√10^3$$...........Answer is yes

Any ODD power will yield always NON-Integer number

Sufficient

(2) $$p = 9 × 10^s$$, where s is a positive integer.

Let s= 1..........$$√9*10$$...........Answer is yes

Let s= 2..........$$√9*10^2$$...........Answer is No

Insufficient

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Director
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Re: Is √p a non-integer? (1) p = r × 10 r, where r is a positive odd in [#permalink]

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25 Sep 2017, 17:17
hazelnut wrote:
Is $$√p$$ a non-integer?

(1) $$p = r × 10^r$$, where r is a positive odd integer.
(2) $$p = 9 × 10^s$$, where s is a positive integer.

Statement 1

In order for the square root of p to be an integer there must be an even number of exponents in the factors of p squared- for example

100 = 5^2 * 2^2
64= 2^6
9= 3^2

suff

Statement 2

Too many possibilities- it is not known whether or not s is odd

insuff

A

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Re: Is √p a non-integer? (1) p = r × 10 r, where r is a positive odd in [#permalink]

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04 Nov 2017, 11:06
1) p = r x10 to power r
--> 1 x 10^1 = 10 --> _/10 is not integer
--> 3 x 10^ 3 = _/3 *1000 not integer
--> 5 x 10^5 = _/5 *100000 not integer
Suff

2) p= 9 x 10^s --> _/9 *10 is s = 1 --> 3_/10 is not Int
is s = 2 then p = 9 *100 --> under root = 3 *10 = integer
Hence, not suff

Ans A

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Is √p a non-integer? (1) p = r × 10 r, where r is a positive odd in [#permalink]

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05 Nov 2017, 11:47
hazelnut wrote:
Is $$√p$$ a non-integer?

(1) $$p = r × 10^r$$, where r is a positive odd integer.
(2) $$p = 9 × 10^s$$, where s is a positive integer.

2 is easier to evaluate, let's start there:
2) $$p = 9 × 10^s$$, if s is odd, then $$\sqrt{p}$$ is a non-integer.
If s is even, then $$\sqrt{p} = \sqrt{(3*3)*(10*10)^\frac{s}{2}} = 3*(10)^\frac{s}{2}$$, an integer.
NS

1) $$\sqrt{p} =\sqrt{r * 10^r} = \sqrt{r * 10 * 10^{(r-1)}}$$
because r is odd, we know $$r-1$$ is even. Therefore $$\sqrt{10^{(r-1)}}$$ is an integer, which leaves us with:
$$\sqrt{r * 10} = \sqrt{r * 2 * 5}$$
because r is odd, there is no way to get the second power of 2 under the radical. The result will always be something$$*\sqrt{2}$$, a non-integer.

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Is √p a non-integer? (1) p = r × 10 r, where r is a positive odd in   [#permalink] 05 Nov 2017, 11:47
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