Really Tough Number Theory Problem

Question

If  N = 2^5 * 3^4 * 5^2 , then

a) Find the total even factors of N.

b) Find the total odd factors of N.

c) Find total prime factors of N.

d) Find total composite factors of n.

e) Find total factors of N which are multiples of 4.

f) Find total factors of N which are multiples of 6.

g) Find total factors of N which are multiples of 8.

h) Find total factors of N which are multiples of 12.

i) Find total factors of N which are multiples of 30.

j) Find total factors of N which will end with at least one trailing zero.

k) Find total factors of N which are perfect squares.

Well, is this within the scope of GMAT?

einstein10 · Answer

go get number of factors of a number is available in OG, so these combinations are also quite possible.
here i propose a very unique way to handle these issues in efficient way:
first write all the factors in following way:
 2^5 * 3^4 * 5^2 =  ( 2^0 * 2^1 * 2^2 * 2^3 * 2^4 * 2^5 )( 3^0 * 3^1 * 3^2 * 3^3 * 3^4 )( 5^0 * 5^1 * 5^2 )

now let's start one by one:

1. number of even factors in above equation will be : count all 2's(except 0 power) * all 3's( along with 0 power)* all 5(along with 0 power)*
  = 5*5*3 = 75

2. number of odd factors: odd factors should not contain 2's. 
so number will be count all 2's(only 0 power) * all 3's( along with 0 power)* all 5(along with 0 power)*
  = 1*5*3 = 15

3. prime factors no need to count it's obvious : 2,3,5 i.e. number of prime factors =3

4. total composite factors , count all 2's(along with 0 power) * all 3's( along with 0 power)* all 5(along with 0 power)*
  = 6*5*3 = 90

5.total number of factors which are multiple of 4's 
count = count all 2's(with more than or equal to power 2) * all 3's( along with 0 power)* all 5(along with 0 power)*
  = 4*5*3 = 60

6. for number of factors 6:
count = count all 2's(excluding 0 power ) * all 3's( excluding 0 power)* all 5(along with 0 power)*
  = 5*4*3 = 60

similarly other counting for other can be done..
your comments are most welcome.. 
please, consider giving kudos if the post is useful thank you..

Hussain15 · Answer

Kudos for your dedicated input!!

But how did you derive the above equation? Is there any kind of combinations involved here?

einstein10 · Answer

it's not formula, it just representation, please not * sign does not mean it's multiplication.. it's simply representations in power of each factor so that visually we can see, how many combinations are possible for counting various ways.

with above representation we can get count any combination of factors in quick way.

actually it simply the permutation of various numbers.. and here basic formula of permutation is involved i.e. total number of ways doing things = m*n

to make more complicated, we can ask what is sum of various combination as you have mentioned. e.g. if we are asked to get the sum of total factors of N which are multiple of 4. this also we can get with the above presentation.

gmat2012 · Answer

i was struggling how to get factors of given number. i will try with this representation gooood post

targetMBA11 · Answer

I havent understood the way u have approached the problem is there any other way to do the same.

mads · Answer

Kudos. Great explanation for findling out factors. Very useful for me.

einstein10 · Answer

please check out this link  https://gmatclub.com/forum/new-set-of-good-ps-85440.html 
problem set 3, Bunuel has explained through formula.. hope it will be useful..

ykaiim · Answer

Is this an assignent? :bebe

einstein10 · Answer

yes, we are all on the board to gmat destination we need more and more assignments to do ..

ykaiim · Answer

There is a formula, if you want to find the total # of factore:

Add 1 to the highest power of each prime factor of a number. For example,
1. 25 = 5^2
Total factors = 1, 5, 25; total 3 factors.
Highest power of 5 is 2, add 1 and it will be = 3.

2. 2500 = 25x100 = 5x5x10x10 = 5x5x5x2x5x2 = 5^4 x 2^2 = 1x2x2x5x5x5x5
All factors = 1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250, 500, 625, 1250, 2500
So, total factors = 15
Now add 1 to individual powers of 5 and 2 i.e., (4+1)x(2+1) = 5x3 = 15.

msand · Answer

find this post very interesting...

Really Tough Number Theory Problem

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