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22 Nov 2016, 22:58
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
44% (01:51) correct
56%
(01:41)
wrong
based on 86
sessions
History
Date
Time
Result
Not Attempted Yet
Is a positive integer x square of an integer?
A) x has exactly 2 different prime factors
B) x has exactly 12 positive divisors including 1 and x.
A) x has exactly 2 different prime factors
B) x has exactly 12 positive divisors including 1 and x.
stonecold
This question is really interesting. we need to find if X is a perfect square.(yes/no)
Lets look at St. 1
X has exactly 2 different prime factors
So we can take the 2 smallest prime factors say 2 and 3, so x is 6, (no)
but if we take the other no like 36 it has 2 different prime factors and is a sq, (6)^2(yes)
We have a yes and a no to st1 so eliminate A and D
lets go to st 2
We know that X has 12 positive factors including one and x
We know the property of perfect squares, i.e they have odd factors.
So we know that x is not a perfect square,
(B) is the correct answer.
General Discussion
23 Nov 2016, 00:21
Kudos
Bookmarks
stonecold
A) \(x=2 \times 3=6\) has extractly 2 prime factors but 6 is not square of an integer.
\(x=2^2 \times 3^2=36\) has extractly 2 prime factors and 36 is square of an integer.
Insufficient.
B) Sufficient.
Let's assume that \(x=p_1^{q_1} \times p_2^{q_2} \times ... \times p_n^{q_n}\), where \(p_1, p_2, ..., p_n\) is distinctive prime factors.
The extractly number of factors (divisors) of \(x\) is \((q_1+1)(q_2+1)...(q_n+1)\)
From (B), we have \((q_1+1)(q_2+1)...(q_n+1)=12\). Let solve for each possible case:
- If \(12=12\), we have \(q_1+1=12 \iff q_1=11\). So \(x=p_1^{11}\), this is not square of an integer (the power is odd).
- If \(12=2 \times 6\), we have \(q_1+1=2\) and \(q_2+1=6\) (No need to care about the order). So \(q_1=1\) and \(q_2=5\). Hence \(x=p_1^1 \times p_2^5\), this is not square of an integer.
- If \(12=3 \times 4\), we have \(q_1+1=3\) and \(q_2+1=4\) (No need to care about the order). So \(q_1=2\) and \(q_2=3\). Hence \(x=p_1^2 \times p_2^3\), this is not square of an integer.
- If \(12=2 \times 2 \times 3\), we have \(q_1+1=2\) and \(q_2+1=2\) and \(q_2+1=3\) (No need to care about the order). So \(q_1=1\) and \(q_2=1\) and \(q_3=2\). Hence \(x=p_1^1 \times p_2^1 \times p_3^2\), this is not square of an integer.
For all cases, we have \(x\) is not a square of an integer. So sufficient.
The answer is B
