PROBLEM 213 — Taking a load off can put a new spin on things as this month’s problem by C. R. Schmidt of Cincinnati demonstrates.
Reknowned sculpture McDuff (pr onounced “McDoof”) Macrae paced the floor agitatedly and muttered under his breath. The final touches on his latest masterpiece were still not right.
The sculpture consists of a spinning globe surrounded by rotating prisms. The prisms rotate at 1 rps, and the globe, at 2 rps. The globe measures 0.46 m in diameter and weighs 80 kg. Assume uniform density of the globe, no windage losses, and frictionless bearings.
If Macrae substitutes another globe having the same size and 10% less mass, how much faster could the second globe rotate if he applies the same amount of energy to start it spinning?
Solution to last month’s problem 212— You know your Ps and Qs if you answered 327654981. Here’s the solution spelled out:
From the clues in the story:
We know that G must be 1 or be divisible by 3. If G 51, then,
which is the largest number possible with the remaining numbers.
H cannot be greater than 1, but we already set G equal to 1. Therefore G is not 1.
The only number not already taken that is divisible by 3 is 9.
Therefore G = 9 and we have:
From (1) we can derive the following relationships:
We know that 10H + I <87, since 9 is already used.
Now, C has to be 8, 7, 5, 4, or 2 (which are the only remaining numbers). Trying all values we find no valid solutions.
The most 20 + C can be is 28.
Therefore, E is 4 or 5, and F must be a whole number, so it is 4 or 8.
Let us take the first case and set E equal to 4. F would then equal 8.
Insert these values in (3) to get
20 + 4 = 20 + C
C cannot equal 4 because we have set E equal to 4. Therefore, E is 5.
F can still be either 4 or 8. If F 58, then (3) becomes:
25 + 4 = 20 + C
Since C cannot equal 9, F = 4 and C equals 7. From here, we can fill in the remaining numbers:
Schnoop saves the day again!