Lesson Relatively prime numbers help to solve problems

Relatively prime numbers help to solve problems

Problem 1

Let  b = the number of boys and  g = the number of girls in the class.

Then we have   = ,  according to the condition.

Let us write it using the common denominator:  = , or, which is the same,  = . 

It implies that 8b = 9g.     (1)

Next, since the numbers 8 and 9 are relatively primes, (1) implies that b is multiple of 9 and g is multiple of 8:

b = 9*n, g = 8*m             (2)

with integer n and m. 

Then you can re-write (1) in the form

8*9n = 9*8m,   or   72n = 72m.

It implies that n = m and, hence, 

b = 9n, g = 8n               (3)

with some integer n.

Now, the number of those students who passed the test is 


 =  =  =  =  = 12n.  (5)

The total number of students in the class is 

b + g = 9n + 8n = 17n.                                                          (6)

Now it is easy to calculate the ratio of those who passed the test to the total number of students in the class. It is (5) divided by (6):

 = .

Answer.  The ratio of those who passed the test to the total number of students in the class is .

Problem 2

The basic equation is

 = .

It implies after cross multiplying

n^2 + 8n = d^2 -d  ====>

n^2 - d^2 = -8n - d  ====>

(n+d)*(n-d) = -8n - d  ====>

(n+d)*(d-n) = 8n + d  ====>

(n+d)*(d-n) = (8n - 8d) + (8d + d)  ====>

(n+d)*(d-n) = -8*(d-n) + 9d  ====>  divide both sides by (d-n)  ====>

n + d = -8 + .   (1)


     By the way (as an aside notice) it follows from the last formula that d > n.


Now, from the very beginning we can assume that our ratio/fraction is  just REDUCED, so n and d have no common factors 
and are relatively prime numbers.

Then "d" and "d-n" in (1) are relatively prime TOO.

Then, since    in (1) is an integer number,  it implies that (d-n) divides 9.


So, the only possible cases are

    a) d-n = 1;
    b) d-n = 3;
    c) d-n = 9.


Case a) d-n = 1 implies (through (1)) that  n+d = -8 + 9d. Then you have this system of two eqns 
                d-n = 1  and  n+d = -8+9d. It has the only solution n=0, d=1, 
                but this solution DOESN't work for global problem.


case b) d-n = 3 implies (through (1) )  n+d = -8 +  = -8+3d. Then you have this system of two eqns  
                d-n = 3 and n+d = -8+3d.  It has the only solution n=2, d=5. 

                Then the fraction is   = , and you can easily check that it works for the global problem.


Case c) d-n = 9 implies (through (1) ) n+d = -8 +  = -8+d.  Then you have this system of two eqns 
                d-n = 9  and  n+d = -8+d.
                but this solution DOESN't work for global problem.