Grades 11-12 Video Solutions 2013-Levels 11&12 Video Solutions 2013 problem21

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Question number 21. How many pairs x, y of integers, with x being less than or equal to y,  exist such that their product equals five times their sum? The equation we would like to solve  is easy to state. We have the product x, y equal to five times the sum of x plus y, and our values  of x and y must be chosen from the integers. Now, it's really an algebra trick to say that we will  let x be equal to u plus five, and y be equal to v plus five, where u and v are integers,  and u will be less than or equal to v, so that x is less than or equal to y. Now, with this tricky  substitution, we see that the product x times y becomes the product quantity u plus five times v  plus five, and here we have on the right-hand side five times the sum u plus v plus 10. We can simplify  a little bit, distribute to obtain u times v plus five u plus five v plus 25 equal to five u plus  five v plus 50, which after simplifying and cancelling gives us the product u times v is equal to 25.  So, as u and v are integers, they must belong to the set of divisors of 25, so plus or minus one,  plus or minus five, plus or minus 25, and so now the pairs u and v where the product is 25 and u is less than v  from among those numbers listed above would be if we choose negative 25, then negative one has to be the  second number. If we choose negative five, then we have no choice but to choose negative five again.  If we choose five, we choose again five, and with one, we choose 25, so that completes the list.  We have other choices, but then we would fail to satisfy the constraint that u is less than v,  and so increasing each coordinate u and v by five gives us the x and y values,  and so we see that there are exactly four choices, so the answer to number 21 is A.

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