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Answer is 25 i.e. D. Can someone please explain how?

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors. For more on number properties check: math-number-theory-88376.html

BACK TO THE ORIGINAL QUESTION: How many factors does 36^2 have? A. 2 B. 8 C. 24 D. 25 E. 26

\(36^2=(2^2*3^2)^2=2^4*3^4\) and according to above it'll have (4+1)(4+1)=25 different positve factros, including 1 and 36^2 itself.

Answer: D.

5 seconds approach: 36^2 is a perfect square. # of factors of perfect square is always odd (as perfect square has even powers of its primes then when adding 1 to each and multiplying them as in above formula you'll get the multiplication of odd numbers which is odd). Only answer choice D is odd thus it must be correct.

Answer: D.

Tips about the perfect square: 1. The number of distinct factors of a perfect square is ALWAYS ODD. 2. The sum of distinct factors of a perfect square is ALWAYS ODD. 3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. 4. Perfect square always has even powers of its prime factors.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Hi Bunuel, can you please clarify when we count multiplies of 1? When do we need to use (1+1)(4+1)(4+1) applicable to this question. Thanks _________________

Don't understand your question. Can you please elaborate?

I mean 36^2=(2^4)*(3^4), the number of factors is (4+1)(4+1)=25 - this is clear. However 36^2 could be (1^1)(2^4)(3^4), which comes to (1+1)(4+1)(4+1)=50. In which cases in GMAT should I count the powers of 1, which doubles the number of factors? Thank you _________________

Don't understand your question. Can you please elaborate?

I mean 36^2=(2^4)*(3^4), the number of factors is (4+1)(4+1)=25 - this is clear. However 36^2 could be (1^1)(2^4)(3^4), which comes to (1+1)(4+1)(4+1)=50. In which cases in GMAT should I count the powers of 1, which doubles the number of factors? Thank you

Understood now. The answer is never.

To find the number of factors of a positive integers we should make its prime factorization. Now, 1 is not a prime, thus you shouldn't write it when making prime factorization of an integer.

1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

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