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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.
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
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