We have discussed how to find the factors of a number and their properties in these two posts:

Writing Factors of an Ugly Number

Today let’s discuss the concept of ‘product of the factors of a number’.

From the two posts above, we know that the factors equidistant from the centre multiply to give the number. We also know that the behaviour is a little different for perfect squares. Let’s take two examples to understand this.

Example 1: Say N = 6

Factors of 6 are 1, 2, 3, 6

1*6 = 6 (first factor * last factor)

2*3 = 6 (second factor and second last factor)

Product of the four factors of 6 is given by 1*6 * 2*3 = 6*6 = 6^2 = [Sqrt(N)]^4

Example 2: Say N = 25 (a perfect square)

Factors of 25 are 1, 5, 25

1*25 = 25 (first factor * last factor)

5*5 = 25 (middle factor multiplied by itself)

Product of the three factors of 25 is given by 1*25 * 5 = 5^3 = [Sqrt(N)]^3

If a number, N, can be expressed as: 2^a * 3^b * 5^c *…

The total number of factors f = (a+1)*(b+1)*(c+1)…

**The product of all factors of N is given by [Sqrt(N)]^f i.e. N^(f/2)**

Let’s look at a couple of questions based on this principle:

Question 1: If the product of all the factors of a positive integer, N, is

2^(18) * 3^(12), how many values can N take?

(A) None

(B) 1

(C) 2

(D) 3

(E) 4

Solution: Since the product of all factors of N has only 2 and 3 as prime factors, N must have two prime factors only: 2 and 3.

Let N = 2^a * 3^b

Given that N^(f/2) = 2^(18) * 3^(12)

(2^a * 3^b)^[(a+1)(b+1)/2] = 2^(18) * 3^(12)

a*(a+1)*(b+1)/2 = 18

b*(a+1)*(b+1)/2 = 12

Dividing the two equations, we get a/b = 3/2

Smallest values: a = 3, b = 2. It satisfies our two equations.

Can we have more values for a and b? Can a = 6 and b = 4? No. Then the product a*(a+1)*(b+1)/2 would be much larger than 18.

So N = 2^3 * 3^2

There is only one such value of N.

Answer (B)

Question 2: If the product of all the factors of a positive integer, N, is 2^9 * 3^9, how many values can N take?

(A) None

(B) 1

(C) 2

(D) 3

(E) 4

Solution: Since the product of all factors of N has only 2 and 3 as prime factors, N must have two prime factors only: 2 and 3.

Let N = 2^a * 3^b

Given that N^(f/2) = 2^(9) * 3^(9)

(2^a * 3^b)^[(a+1)(b+1)/2] = 2^(9) * 3^(9)

a*(a+1)*(b+1)/2 = 9

b*(a+1)*(b+1)/2 = 9

Dividing the two equations, we get a/b = 1/1

Smallest values: a = 1, b = 1 – Does not satisfy our equation

Next set of values: a = 2, b = 2 – Satisfies our equations

All larger values will not satisfy our equations.

Answer (B)

Note that we can easily use hit and trial in these questions without actually working through the equations.

This is how we will do it:

N^(f/2) = 2^(18) * 3^(12)

Case 1: Assume values of f/2 from common factors of 18 and 12 – say 2

[2^9 * 3^6]^2

Can f/2 = 2 i.e. can f = 4?

If N = 2^9 * 3^6, total number of factors f = (9+1)*(6+1) = 70

This doesn’t work.

Case 2: Assume f/2 is 6

[2^3 * 3^2]^6

Can f/2 = 6 i.e. can f = 12?

If N = 2^3 * 3^2, total number of factors f = (3+1)*(2+1) = 12

This works.

The reason hit and trial isn’t a bad idea is that there will be only one such set of values. If we can quickly find it, we are done.

Why should we then bother to find it at all. Shouldn’t we just answer with option ‘B’ in both cases? Think of a case in which the product of all factors is given as 2^(16) * 3^(14). Will there be any value of N in such a case?

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*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*