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

Question

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.

argerniki · Accepted Answer

Question: is p a perfect square?

(1)  p = r * 10^r

p = r * 10^r , which also means that  p = r * (2*5)^r = r * 2^r * 5^r 
In order to be a perfect square, p has to consist out prime factors with only even exponents

As we know that r is odd, the exponents of 2 and 5 are odd as well.
Therefor, r needs to have at least 2 and 5 as prime factors, so that the exponents of 2 and 5 will become even (remember: odd number + 1 = even number).

r can not have 2 as a prime factor as it is an odd number.

Therefor p CAN'T be an an perfect square.

SUFFICIENT

(2)  p = 9 * 10^s

p = 9 * 10^s = 3^2 * 10^s

3 already has an even exponent, so that 10 needs an even exponent as well.
Therefor p is a perfect square when s = even.
When s = odd, p IS NOT a perfect square.

UNSUFF

There the answer ist A imo

TestPrepUnlimited · Answer

We can also ask "is  √p  an integer?" or  "is  p  a perfect square?"

Statement 1:

All squares can be broken down to even powers of prime factors, or a product of squares. 
Split p into squares,  r * 10^r = r * 10 * 10^{r-1} .  10^{r-1}  is a square since  r - 1  is even. Then the concern is whether the rest of the factors,  r * 10  can be a square.  r  needs to have a factor of 2 since 10 has a factor of 2, to complete the square. Thus  r  has to be even if we want to complete the square, then there is no odd  r  that can make  r * 10  and  r * 10^r  a square. Sufficient.

Statement2:

Again focus on breaking p into smaller squares. 9 is already a square, we are concerned about  10^s  now. We only know  s  is an integer, any even  s  would make  10^s  a square but any odd  s  wouldn't. So this is insufficient.

Ans: A

Farina · Answer

Why perfect square? The question stem is asking about integer value and in the given info (1) and (2) the number 10 is given that will give answer in integer form in both cases, so why not D?

fmcgee1 · Answer

I'm finding this really hard to understand.

So  10^{r-1}  is a perfect square because the power is even, right? But then when we look at the other two numbers, 10 and r, the prime factors of 10 are  5^{1}  and  2^{1} .... which would mean that r would have to also have an odd power of both 5 and 2 in order to make the whole term a square. And we have no way of knowing what r is or how many powers of 5 or 2 it could have.... so how is the answer A?

Basshead · Answer

Is  √p  a non-integer? = Is p a perfect square?

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

Prime factorization =  r * 5^r * 2^r 
Since we are told r is a positive odd integer, we can make the conclusion that p is not a perfect square. To be a perfect square each exponent needs to be  even .

We can also plug in numbers. Let r = 1

1 * 10 = 10
10 = not a perfect square.

Sufficient.

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

3^2 * 10^s

If s is even, then p is a perfect square. Insufficient.

bumpbot · Answer

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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

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