Odd numbers with perfect cubes.
Number Theory
Level
2
What is the probability that a number chosen b/w 1 and 1000(both inclusive) has odd number of divisors given that it is a perfect cube ?
3/10
2/5
1/5
1/2
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1 solution
There are 10 perfect cubes between 1 and 1000, inclusive. Each of these numbers can be expressed in the form x^3. If x is a prime between 1 and 10, inclusive, then it will have 3+1=4 factors, which is an even number. If x is a non-square, even number, then it will also have an even number of factors since it will be in the form x^3 * y^3 ... However, if x is a square, then it will have an odd number of factors since it will have the form x^3=(x/2)^6, which will have 7 factors. There are three perfect squares between 1 and 10, inclusive. Thus, the probability is 3/10.