divisors

Question

find all the numbers less than 100 with exactly 8 divisors.

AkamaiBrah · Answer

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.

vcbabu · Answer

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 .

vesam · Answer

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

divisors

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