Our regular commentator Manuel Amorós has simplified the solution to the ox problem, posed last week, in an ingenious way that would have undoubtedly pleased Newton himself:
“I will propose an algebraic procedure that does not require too much thought. Let us suppose that the initial height of the grass is h and the weekly growth rate is v. We can then establish the following proportions.
10/3(h + 4v)—————12 oxen———————4 weeks
10(h + 9v)——————21 oxen———————9 weeks
24(h + 18v)——————x oxen——————–18 weeks
From the first two we obtain the relationship h = 12v
Finally we can clear x = 36
Using the same procedure, I think that in the second case 100 is the solution.”
Along the same lines of saving mental energy, Susana Luu says:
“The long hand makes one revolution per hour and the short hand makes one revolution every 12 hours. Suppose 5 minutes have passed since 12 o’clock. The long hand will be at 5 minutes, but the short hand is no longer at 0 minutes, but slightly ahead, so this is not a valid result. If we allow another “almost 5 minutes” (just under 5 minutes) to pass, the long hand will be almost at 2, but the short hand will now be behind. Again this is not a valid result, but assuming continuity, the short hand, going from ahead to the valid result to behind to the valid result, will at some point show a valid result. It is easy to calculate such a time explicitly, but it is not necessary to know that it exists. The same reasoning applies for all subsequent 5-minute intervals, with the sole exception of the two daily intervals where we arrive at 12 o’clock. Thus there will be as many valid moments per day as there are 5-minute intervals in 24 hours minus 2. Total: 286″.
That is, 143 every 12 hours. And Manuel Amorós clarifies:
“Each solution position is formed by the position of two needles, but the fact that the needles coincide does not imply that they are two solutions in one, it is ONE solution like any other. What does happen is that there is another solution that also results in the same position of the needles at the initial point, so in effect there are 43 solutions.” What is that other solution?
For his part, Francisco Montesinos considers that the movement of the minute hand is not continuous (electric clock), but moves in leaps of one minute (mechanical clock), which introduces an interesting variant (see comments from last week).
Since we have solved (or at least tried to) several algebraic problems lately, it is a good time to remember that, when moving from elementary arithmetic to algebra, the usual “four operations” are usually no longer sufficient. Just as an addition with the added end repeated is a multiplication, a multiplication with the factor repeated is a power. And just as addition has its opposite operation in subtraction and multiplication in division, exponentiation has its opposite in root extraction, that is, obtaining square roots, cubic roots, etc. That’s six operations already… What’s the seventh? (Even if you don’t remember it, you can deduce it by extrapolating from the above).
While you think about it, here are three little problems regarding the sixth operation:
Which is greater, the fifth root of 5 or the square root of 2?
Which is greater, the fourth root of 4 or the seventh root of 7?
Which is greater, √7 + √10 or √3 + √19?
In all three cases, it is a matter of finding the solution mentally (or, if you cannot do without paper and pencil, by performing only very simple arithmetic operations).
#Musketeers
