how many factors does a number have?

[gobaudd]

On page 58 of the number properties guide in the answers to chapter 4 the answer to #11 mentions that thee are (4 + 1)(1 + 1)=10 different factors of 2^4 * 3. How is it calculating this? I t refers to chapter 1 but I don't remember any such formula.

also, why is 3 a factor? it says "because there are 3 consecutive even numbers" but I don't understand why it is true. on page 53 it says the product of k consecutive integers is always divisible by k! which i understand but does that extend to any set of evenly spaced numbers?

Thanks

[gokul_nair1984]

On page 58 of the number properties guide in the answers to chapter 4 the answer to #11 mentions that thee are (4 + 1)(1 + 1)=10 different factors of 2^4 * 3. How is it calculating this? I t refers to chapter 1 but I don't remember any such formula.

I do not have the MGMAT book but I can help you with this.

To find the total number of factors for any number , express the number in terms of its prime factors.
eg: 48=2*2*2*2*3

We have 2 prime factors here, viz. 2 and 3.

Therefore, 48 =(2^4)*(3^1).

As a general rule the total number of factors will be
(4+1)*(1+1)=10

Generally speaking if a number, n can be expressed as :
n=(a^x)*(b^y), where a and b are prime numbers, then total number of factors is given by (x+1)*(y+1)(This includes 1 and the number itself)

Another example 100=(5^2)*(2^2).
Therefore total number of factors of 100 will be (2+1)*(2+1)=9

also, why is 3 a factor? it says "because there are 3 consecutive even numbers" but I don't understand why it is true.

Take any three consecutive even numbers. Let's say 2,4,6. It's product has to be divisible by 3. Among any 3 even consecutive numbers , there will always be a number that has 3 as its factor. Analyzing that the factors of 3 are spaced out at an interval of 3(3,6,9,....) and the factors of 2 are spaced out at an interval of 2(2,4,6) might be helpful in this regard.Furhtermore, every third multiple of 2 will be divisible by 3 as 2*3=6

Hope I could help.

[chitrangada.maitra]

Also, the reason they refer you to chapter 1 is because the concept is explained in detail in chapter 1, pg 15, under 'factors and multiples'.

n page 53 it says the product of k consecutive integers is always divisible by k! which i understand but does that extend to any set of evenly spaced numbers?

This concept does not apply to all evenly spaced sets sets
e.g product of evenly spaced set of 3 numbers: (2,5,8) is not divisible by 3

[tim]

i hope these explanations help. let us know if you need any further explanations..

[gobaudd]

thanks i actually found the answer on page 130 in the book so i think the reference in chapter 4 to chapter is incorrect and should instead refer to page 130.

my other post inequality-optimization-exponent-question-t11024.html still needs answered though.

thanks

[tim]

we'll get to it eventually..