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divisors

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divisors [#permalink] New post 28 May 2009, 08:52
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find all the numbers less than 100 with exactly 8 divisors.
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Re: divisors [#permalink] New post 30 May 2009, 00:54
vcbabu wrote:
find all the numbers less than 100 with exactly 8 divisors.


This is an interesting problem. Let's approach this logically. Here is a big hint:

because there need to be exactly 8 divisors, we need one of two possible factorizations:

A^1*B^1*C^1 < 100 where A, B, and C are prime. (2 x 2 x 2 combinations of exponents -- don't forget the zeroth power!)

or A^3 * B^1 < 100 where A, and B are prime. (4 x 2 combinations of exponents).

Note that there is no other way to factor this where you will get exactly 8 (well, technically there is A^7, but 2^7, which is the smallest number, = 128 so that won't work).

With a little bit of common sense and cut and try, we can narrow this down quickly since there are only so many possible combinations that yield a result < 100. I'm going to leave the rest of the solution for you to solve.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Re: divisors [#permalink] New post 30 May 2009, 12:35
no of divisors of a number n is (a+1) * (b+1)*(c+1) if n is x^a * y^b*z^c where x,y,z are primes .

hence

8 = 2*2*2 =(1+1)*(1+1)*(1+1)

and the number shall be <100 .

so 2*3*5 ; 2*3*7, --- 5 combinations .

also
it could be 8= 4*2 =(3+1) * (1+1) = 2^3*3; 2^3*5;2^3*7 ; 2^3*11;3^3*2 so 5 combinations .

total 10 .


lokking for any faster way of solving the problem .
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Re: divisors [#permalink] New post 31 May 2009, 21:47
Let us take it with equation methid.
For total number of factors 'n', n = (p+1) (q+1) (r+1)....
So the actual number formed will be a^p*b^q*c^r..., where a, b, c,... are prime numbers

We have n = 8, which can be splitted as follows
8 = 2*2*2 or 4*2 or 1*8
(i)
=> 8 = (1+1)(1+1)*(1+1) --> in (p+1) (q+1) (r+1).... format
No we have (p,q,r) = (1,1,1)

Substitute them with different prime number combinations
1. 2^1 * 3^1 * 5^1 = 30
2. 2^1 * 3^1 * 7^1 = 42
3. 2^1 * 3^1 * 11^1 = 66
4. 2^1 * 3^1 * 13^1 = 78
5. 2^1 * 5^1 * 7^1 = 70
Now we have 5 numbers with this 2*2*2

(ii) 8 = 4*2
=> 8 = (3+1)*(1+1)
Now we have (p,q) = (3,1)
Substitute them with different prime number combinations
6. 2^3 * 3^1 = 24
7. 2^3 * 5^1 = 40
8. 2^3 * 7^1 = 56
9. 2^3 * 11^1 = 88
10. 3^3 * 2^1 = 54
Now we have 5 numbers with this 4*2

(iii) 8 = 1* 8
=> 8 = (0+1)*(7+1)
Now we have (p,q) = (0,7)
Minumum prime number is 2, but 2^7 = 132 > 100, so out of choice,

Finally we have 10 numbers having exactly 8 factorials from 1 to 100
{24, 30, 40, 42, 54, 56, 66, 70, 78, 80}

Answer is 10.

Thanks
Re: divisors   [#permalink] 31 May 2009, 21:47
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