Lat "a" be a positive numbere which is greater than 1 , say N=a^pb^qr^swherw a,b,c are diffrent primes and ?

Question

the number of ways in which N can be resolved is

N=1\2 (p+) (q+1) (s+1)

can any one explain it very clearly ,in few sentences ,

be swift!!!!

be fast!!!!

Answer 1

If I can clarify, this appears to be the problem, where, if

n = a^p * b^q * c^r (being the prime factorisation of n),

where a, b, c are different primes and p, q, r are integers

(not necessarily different), then the number of distinct

divisors of n is given by the number-theoretic function

T(n) = (p + 1)(q + 1)(r + 1).

To see how simply this works, let a = 2, b = 3, c = 5

and let p = 2, q = 2, r = 2.

Thus, N = 2^2 * 3^2 * 5^2 = 2 * 2 * 3 * 3 * 5 * 5

Now all we have to do is find all the different combinations

of factors, 1 at a time, 2 at a time, etc., up to 6 at a time.

Here they are, all 27 of them, i.e. (2 + 1)(2 + 1)(2 + 1).

1 : 1 (1 divides into all integers)

2 : 2

3 : 3

5 : 5

4 : 2 * 2

6 : 2 * 3

10: 2 * 5

9 : 3 * 3

15: 3 * 5

25: 5 * 5

12 : 2 * 2 * 3

20 : 2 * 2 * 5

18 : 2 * 3 * 3

30 : 2 * 3 * 5

50 : 2 * 5 * 5

45 : 3 * 3 * 5

75 : 3 * 5 * 5

36 : 2 * 2 * 3 * 3

60 : 2 * 2 * 3 * 5

100: 2 * 2 * 5 * 5

90 : 2 * 3 * 3 * 5

150: 2 * 3 * 5 * 5

225: 3 * 3 * 5 * 5

180 : 2 * 2 * 3 * 3 * 5

300 : 2 * 2 * 3 * 5 * 5

450 : 2 * 3 * 3 * 5 * 5

900 : 2 * 2 * 3 * 3 * 5 * 5

So, each factor is of the form a^x * b^y * c^z,

where x ≤ p, y ≤ q and z ≤ r.

Because p = 2, then x can only be 0, 1 or 2 ( 3 values).

Likewise, y and z can also only be one of these 3 values.

So with n = a^p * b^q * c^r, a^p could be 2^0, 2^1 or 2^2,

b^q could be 3^0, 3^1 or 3^2 and c^r could be 5^0, 5^1 or 5^2.

3 ways for first position times 3 ways for second position

times 3 ways for third position = 27 combinations.

The number of ways for each integer p, q, r is thus p+1,

q+1 and r+1, the "+1" being for the prime raised to zero,

in addition to the actual values of p, q and r.