Economist wrote:
Hi Bunuel, would appreciate if you can explain the solutions for 3,5 and 9.
3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
Answer: D.
First of all you should know the formula counting the number of distinc factors of an integer:
You have to write the number as the product of primes as
a^p*b^q*c^r, where a, b, and c are prime factors and p,q, and r are their powers.
The number of factors the number contains will be expressed by the formula (p+1)(q+1)(r+1).Let's take an example for clear understanding:Find the number of all (distinct) factors of 1435:
1. 1435 can be expressed as 5^1*17^1*19^1
2. total number of factors of 1435 including 1 and 1435 itself is (1+1)*(1+1)*(1+1)=2*2*2=8 factors.
OR
Distinct factors of 18=2*3^2 --> (1+1)*(2+1)=6. Lets check: factors of 18 are: 1, 2, 3, 6, 9 an 18 itself. Total 6.
Back to our question:
How many numbers that are not divisible by 6 divide evenly into 264,600?
264,600=2^3*3^3*5^2*7^2
We should find the factor which contain no 2 and 3 together, so not to be divisible by 6.
Clearly, the factors which contain only 2,5,7 and 3,5,7 won't be divisible by 6. So how many such factors are there?
2^3*5^2*7^2 --> (3+1)*(2+1)*(2+1)=36 (the product of powers of 2, 5,and 7 added 1)
3^3*5^2*7^2 --> (3+1)*(2+1)*(2+1)=36 (the product of powers of 2, 5, and 7 added 1)
So 36+36=72. BUT this number contains duplicates:
For example: 2^3*5^2*7^2--> (3+1)*(2+1)*(2+1)=36 This 36 contains the factors when the power of 2 is 0 (2^0=1)--> 2^0*5^2*7^2 giving us only the factors which contain 5-s and/or 7-s. (5*7=35, 5*7^2=245, 5^2*7=175, 5*7^0=5, 5^0*7=7....) number of such factors are (2+1)*(2+1)=9 (the product of powers of 5 and 7 added 1).
And the same factors are counted in formula 3^3*5^2*7^2 --> (3+1)*(2+1)*(2+1)=36: when power of 3 is 0 (3^0=1). --> 5*7=35, 5*7^2=245, 5^2*7=175, 5*7^0=5, 5^0*7=7.... such factors are (2+1)*(2+1)=9. (the product of powers of 5 and 7 added 1).
So we should subtract this 9 duplicated factors from 72 --> 72-9=63. Is the correct answer.
The problem can be solved from another side:
264,600=2^3*3^3*5^2*7^2 # of factors= (3+1)(3+1)(2+1)(2+1)=144. So our number contains 144 distinct factors. # of factors which contain 2 and 3 is 3*3=9 (2*3, 2^2*3, 2^3*3, 2*3^2, 2^2*3^2, 2^3*3^2, 2*3^3, 2^2*3^3, 2^3*3^3 total 9) multiplied by (2+1)*(2+1)=9 (powers of 5 and 7 plus 1) --> 9*9=81 ---> 144-81=63.
Hope now it's clear.