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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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New post 23 Mar 2017, 18:11
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Question Stats:

67% (01:30) correct 33% (01:51) wrong based on 146 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.

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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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New post 23 Mar 2017, 19:10
avinsethi wrote:
A?

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

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New post 23 Mar 2017, 19:20
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

Answer 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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New post 25 Sep 2017, 18:17
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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New post 04 Nov 2017, 12:06
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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New post 05 Nov 2017, 12:47
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.

The answer is A.
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Is √p a non-integer? (1) p = r × 10 r, where r is a positive odd in &nbs [#permalink] 05 Nov 2017, 12:47
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