Latest TCS Aptitude Question SOLUTION: A number has exactly three prime factors.125 factors of this number are perfect squares. 25 number of factors are perfect cube.What is the minimum number of factors possible for th

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A number has exactly three prime factors.125 factors of this number are perfect squares. 25 number of factors are perfect cube.What is the minimum number of factors possible for this number....
1. 343
2. 512
3. 1000
4.729

Asked In TCS VISWANATH (10 years ago)

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PARAG KAMRA  (10 years ago)

We know that the total factors of a number N = ap.bq.cr ....
Now the total factors which are perfect squares of a number N = ([p2]+1).([q2]+1).([r2]+1)....
where [x] is greatest intezer less than that of x.
Given ([p2]+1).([q2]+1).([r2]+1).... = 125
So [p2]+1 = 5; [q2]+1 = 5; [p2]+1 = 5
[p2] = 4 ⇒ p = 8 or 9, similarly q = 8 or 9, r = 8 or 9
Given that 27 factors of this number are perfect cubes
so ([p3]+1).([q3]+1).([r3]+1).... = 27
So [p3]+1 = 3 ⇒ = [p3] = 2
⇒ p = 6, 7, 8
By combining we know that p = q = r = 8
So the given number should be in the format = a8.b8.c8 ....
Number of factors of this number = (8+1).(8+1).(8+1) = 729

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Kavisha  (5 years ago)

First, we notice we have a number N with three prime factors, so N=pa1pb2pc3.

Now, the number of factors of N equals (a+1)(b+1)(c+1). Of these factors, (⌊a2⌋+1)(⌊b2⌋+1)(⌊c2⌋+1) are perfect squares, and (⌊a3⌋+1)(⌊b3⌋+1)(⌊c3⌋+1) are perfect cubes.

Now notice that both 27 and 125 ar third powers, implying that a, b and c are probably equal (this is provable by using their prime factorizations).

We're now looking for a number a which satisfies (⌊a2⌋+1)=5 and (⌊a3⌋+1)=3.

By using brute force we determine that a=b=c=8.

Plugging this into the original equation gives us that N has 729 prime factors.

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