parheliubdr

2023-02-20

How many ordered pairs of positive integers (m,n) are there such that the least common multiple of m and n is 2^3 7^4 13^13?

Camuccinirk84

Beginner2023-02-21Added 4 answers

Both m and n factors of 2^3 7^4 13^13.

So m=2^a_1 7^b_1 13^_c1 and n=2^a_2 7^a_2 13^a_2

For some non-negative integers a1,b1,c1,a2,b2,c2.

2^3 y^4 13^13 is the least common multiple.

max {a1,a2}=3

max {b1,b2}=4

max {c1,c2}=13

{a1a2} can be equal to

(0,3),(1,3),(2,3),(3,3),(3,2),(3,1) or 3,0) a) total of 7 choices.

(b1,b2)=4 for {b1,b2} we have 2*4+1=9 Choices.

Maximum {c1,c2}=13, for {c1,c2} we have 2*13+1=27 choices.

And the number of ordered pairs (m,n)=7*9*27=1701

So m=2^a_1 7^b_1 13^_c1 and n=2^a_2 7^a_2 13^a_2

For some non-negative integers a1,b1,c1,a2,b2,c2.

2^3 y^4 13^13 is the least common multiple.

max {a1,a2}=3

max {b1,b2}=4

max {c1,c2}=13

{a1a2} can be equal to

(0,3),(1,3),(2,3),(3,3),(3,2),(3,1) or 3,0) a) total of 7 choices.

(b1,b2)=4 for {b1,b2} we have 2*4+1=9 Choices.

Maximum {c1,c2}=13, for {c1,c2} we have 2*13+1=27 choices.

And the number of ordered pairs (m,n)=7*9*27=1701