Grades 11-12 Video Solutions 2013-Levels 11&12 Video Solutions 2013 problem19

Video Transcription

Question number 19, how many pairs x, y of positive integers,  satisfy the equation x squared times y cubed is equal to 6 to the power of 12?  We will need the prime factorization of both x and y.  So let's factor the right-hand side.  We have 6 to the power of 12 factors as 2 to the power of 12 times 3 to the power of 12.  And so somehow we have to recombine the factors and the exponents, so there is an x to the power of 2 and a y to the power of 3, both positive integers.  So we will set x equal to 2 to some power times 3 to a power  and y will be similarly 2 to a power times 3 to a power.  So then x squared times y cubed, we can multiply those powers of 2 and 3 on combined basis.  So we will have 2 to the power a, 3 to the power b,  that's x squared times 2 to the power c, 3 to the power d, which is y cubed, and combining bases we obtain 2 to the power 2a plus 3c and 3 to the power 2b plus 3d.  Now, this was supposed to be equal to 2 to the power 12 times 3 to the power 12.  So what we are doing is looking at the equations 12 equal to 2a plus 3c and 12 equal to 2b plus 3d,  where a, b, c, d are positive integers.  And what we are writing is the factorization of 6 to the power 12 into x squared and y cubed,  where x we are associating with the pair a, c,  and y we are associating with the pair b, d. So the number of ways in which we can uniquely choose  such pairs will be the number of positive integers, x, y, satisfying the constraints of the problem. So let's solve one equation here.  As they are the same, we have 12 is equal to 2a plus 3c,  12 minus 3c divided by 2 is a. So we can say that c has to be a positive integer, so it must be a. So the choices are c is equal to 0, in which case a is equal  to 6, c is equal to 2, in which case a is equal to 3, and finally c can be 4, and then a would be 0.  So the pair a, c corresponding to x and likewise the pair b, d corresponding to y must be chosen to the set of pairs  where we have 6, 0, 3, 2, and 0, 4.  And now we have to choose them in a fashion that will produce x and y. So we choose with repetition  and we have 3 times 3 choices,  which gives us 9 possible pairs of integers.  And since that is not a valid choice here, we conclude that the answer to the problem is e.

Video Summary

The video transcript discusses solving a mathematical problem involving finding the number of pairs \(x, y\) of positive integers that satisfy the equation \(x^2 y^3 = 6^{12}\). The solution involves prime factorizing \(6^{12}\) into \(2^{12} \times 3^{12}\) and assigning powers \(a, b, c, d\) to the factors of \(x\) and \(y\). By equating powers, two equations \(12 = 2a + 3c\) and \(12 = 2b + 3d\) are solved for positive integer solutions. Ultimately, there are 9 valid pair combinations of \(x, y\), making this the final solution.

Keywords

mathematical problem

positive integers

prime factorization

equation solutions

pair combinations