Question number 28. Consider positive integers a, b, and c that satisfy the condition that a squared is equal to 2b cubed is equal to 3c to the power 5. The smallest possible number of all positive divisors of this product, a, b, and c, including one in the product itself, is then which number? We have to factor a, b, and c somehow and then multiply them together. We notice that 2 and 3 are divisors of a, b, and c. We can write, in this case, A as the product of 2 to some power, let's call it k1, times 3 to some power, let's call it k2, and times some number a1, yet to be determined. B is similarly factored as 2 to some power, let's call it l1, 3, l2, times some number b1, yet to be determined, and C likewise will be 2 to some power, let's call it m1, 3 to the power m2, and B, and the last number would be aC1. So then the product, a, b, c, is given by 2 to the power k1 plus l1 plus m1, 3 to the power k2 plus l2 plus m2, times a1 times b1 times c1. And to minimize the number of factors here, we will have to choose this to be 1, set equal to 1, we can choose these numbers as we like, and the exponents here we have to minimize, and that will give us the minimum number of divisors. So let's look at the exponents again, we haven't used our equation yet, so we can write that a squared is equal to 2k1, 3k2, a1, all that to the second power, that's equal to 2 times b cubed, so 2l1, 3l2, b1 to the third power, and that's equal to 3 times c to the power fifth, 2m1 times 3m2 times c1, all to the power 5, and then equate exponents. So the powers of 2, we add up the exponents in each equation, we have 2k1 from the first equation, that's equal to 3l1 plus, we have an extra power of 2, and 5m1 in the third equation, likewise for the powers of 3, we have 2k1 equal to 3l2, and that's equal to 5m2 with an extra power of 3, so 5m2 plus 1, and we have to find the minimal solutions here, so the smallest values of k, l, and m in each case, that give us a solution, so we will set here in the top equation, m equal to 2, so that we have 10 here, we need l1 to be equal to 3, and k equal to 5, in the second equation we choose m2 to be 1, so that we have a 6 everywhere, if l2 is 2, and k1 is 3, so these would be the minimal solutions, and then we see that our product with this choice, we have a, b, c is equal to 2 to the power k1 plus l1 plus m1 would give us a power of 10, and times 3 to the power k2l2 plus m2 would give us a power of 6, and then we have a1 times b1 times c1, which we choose to be 1, so in that case we have the product of the exponents plus 1 would be the number of the divisors, 10 plus 1 times 6 plus 1, the plus 1 accounts for the number itself, and 1 if we choose exponent 0, so that is 11 times 7 or 77 as the number of divisors, and the answer is then d.

positive integers

divisors

exponents

factorization

minimal solutions