How do you find a and b?

i) a is such that has 12 divisors
ii) b has 18 divisors and
iii) gcd (a,b)=45

2 Answers
Feb 26, 2018

a=90a=90
b=171b=171

Explanation:

a=45x_1a=45x1

b=45x_2b=45x2

Divisors of 45 -> 1xx45, color(white)("d")3xx15,color(white)("d")5xx9451×45,d3×15,d5×9 count of 6

Given that a->12a12 divisors then 2xx45->2xx"6 divisors"2×452×6 divisors

Given that b->18b18 divisors then 3xx45->3xx"6 divisors"3×453×6 divisors

3xx45= 1713×45=171
2xx45=902×45=90

Tony B

Tony B

Mar 1, 2018

Another approach.

Explanation:

Given a number N = prod_(k=1)^n p_k^(alpha_k)N=nk=1pαkk

with p_kpk being the primes and alpha_kαk their multiplicity we have

prod_(k=1)^n (alpha_k+1)nk=1(αk+1) divisors for NN

Assuming now that aa and bb are composed by the same prime factors 33 and 55 we have

a = 3^2 xx 5^(alpha_1)a=32×5α1
b = 3^(alpha_2) xx 5b=3α2×5

The number of factors for aa and bb are

{(a->(2+1)(alpha_1+1)= 12),(b->(alpha_2+1)(1+1)=18):}

now solving for alpha_1, alpha_2 we have

alpha_1 = 3 and alpha_2 = 8

then

a = 3^2 xx 5^3 and
b = 3^8 xx 5

a factors are

{1, 3, 5, 9, 15, 25, 45, 75, 125, 225, 375, 1125}

b factors are

{1, 3, 5, 9, 15, 27, 45, 81, 135, 243, 405, 729, 1215, 2187, 3645, 6561, 10935, 32805}