Question

Use the Multiplication Rule to find the number of positive divisors of 20!. Include a procedure that “builds” such divisors

Answer 1

The 20! can be written as

20 = 20 * 19 * 18 * 17 *16 * 15 * 14 *13* 12*11 * 10 * 9 * 8 * 7 * 6 * 5 * 4* 3 *2 *1

20!=(4*5) * 19 * (2* 9) * 17 * ( 2*8) * (3*5) * (2*7) * (2*6) * 11 * (2 *5) * (3*3) * (2*4) * 7 * (2*3) * 5 *(2*2) *3 *2 *1

20! = (2² *5) *19 * (2 * 3²) *17 * (2⁴) * (3*5) *(2*7) * (2²*3) *11 *(2*5) * (3²) *(2³) *7 * (2*3) * 5 *(2²) *3 *2 *1

20! = 2¹⁸*3⁸*5⁴*7²*11*13*17*19

Rule : If number n can be written as

n = p₁^a * p₂^b *...........*p_i^k.

where p_1, p₂ ,.......p_i are the primes and a, b, ...........k are the powers.

Then number of positive divisors of n are

d(n) = (a+1) * (b+1) *..........*(k+1).

Here n = 20!

p₁=2, p₂ = 3, p₃=5, p₄=7, p₅ =11, p₆ =13, p₇=17, p₈ = 19

a= 18, b= 8, c= 4, d=2, e=1,f=1,g=1,h=1

Hence number of positive divisors of 20! is

d(20!) = (18+1) * (8+1) * (4+1) * (2+1)*(1+1)*(1+1)*(1+1)*(1+1)

d(20!) = 19 * 9 * 5 * 3 *2*2*2*2

d(20!) = 41040

Number of positive divisors of 20! = 41040.