Lesson Relatively prime numbers help to solve problems
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<H2>Relatively prime numbers help to solve problems</H2> <H3>Problem 1</H3>A group of boys and girls sit a test. Exactly 2/3 of the boys and 3/4 of the girls pass the test. If an equal number of boys and girls passed the test, what fraction of the entire group passed the test? <B>Solution</B> <pre> Let b = the number of boys and g = the number of girls in the class. Then we have {{{(2/3)*b}}} = {{{(3/4)*g}}}, according to the condition. Let us write it using the common denominator: {{{(8/12)*b}}} = {{{(9/12)*g}}}, or, which is the same, {{{(8b)/12}}} = {{{(9g)/12}}}. It implies that 8b = 9g. (1) Next, since the numbers 8 and 9 are <U>relatively primes</U>, (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 {{{(2/3)*b + (3/4)*g}}} = {{{(8b)/12 + (9g)/12}}} = {{{(8*9n)/12 + (9*8n)/12}}} = {{{(72n)/12 + (72*n)/12}}} = {{{(144/12)*n}}} = 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): {{{(12n)/(17n)}}} = {{{12/17}}}. <U>Answer</U>. The ratio of those who passed the test to the total number of students in the class is {{{12/17}}}. </pre> <H3>Problem 2</H3>The numerator of a fraction is increased by 8 and the denominator is decreased by 1, the resulting fraction is the reciprocal of the original fraction. What is the original fraction? <B>Solution</B> <pre> The basic equation is {{{(n+8)/(d-1)}}} = {{{d/n}}}. 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 + {{{(9d)/(d-n)}}}. (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 {{{(9d)/(d-n)}}} 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 + {{{(9*d)/3}}} = -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 {{{n/d}}} = {{{2/5}}}, and you can easily check that it works for the global problem. Case c) d-n = 9 implies (through (1) ) n+d = -8 + {{{(9*d)/9}}} = -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. </pre> Thus the only solution for global problem is n = 2, d = 5 with the fraction 2/5 = {{{2/5}}}. 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