A New Favorite Puzzle




32 thoughts on “A New Favorite Puzzle

  1. Once again, you have made my day! I did not begin to think to “power” my way out of this (or should I say “exponentiate” –which I believe is still legal in the state of North Carolina!)

  2. classic maths degree student (or math major for the Yanks out there) I turned straight to logarithms. I much prefer this method though, very nice

  3. I didn’t notice at first the choice of power (30th) was conveniently chosen to be product of the two powers (10th and 3rd) of the two original terms. In fact the final slide suggested to me that the 30th power should be used, again, to compare fourth root of two and tenth root of six.

    Maybe making this “common multiple” thing more explicit would be helpful to others?

      1. Yes, if 12 perfect fifths perfectly matched 7 octaves, then consistent musical scales could be constructed with desirable intervals in any key. Because the fifths don’t line up with the octaves, various “temparaments” have been used, including the “well-tempered” approach promoted by Bach.

        1. My music theory is a bit rusty, remind me why they don’t match up? And what do you mean “consistent musical scales with desirable intervals”?

        2. Each perfect fifth represents a 3:2 ratio of frequencies. Twelve of these would be a ratio of (3/2)^12 to 1, which is 129.75:1 and not an exact match for 128:1, the frequency ratio for seven octaves (2^7). If the musical math worked out, we could have a keyboard instrument for which every key had perfect fifths, perfect fourths (4:3) and other intervals as ratios of integers. The appeal of these intervals was noted as far back as Pythagoras. A “just” tuning can provide these intervals for one or two keys but other keys will sound terrible.

      1. Yeah, I’m with you – the key fact is that raising both sides of an inequality to a power preserves the inequality. (Assuming all numbers are positive.)

        Interestingly, a few people have suggested that this property fails if the numbers are between 0 and 1, but in fact this approach handles such a situation without a hitch!

  4. do the “to the power of 30” is still complicated for hand calculation. To the power of 10 is enough and then comparing 10=8*(10/8) and 2^(10/3)=(2^3)*[2^(1/3)]=8*2^1/3; therefore just by comparing 1.25 and 2^1/3

    1. Comparing 10^(30/10) to 2^(30/3) isn’t complicated to do by hand. I’d rather compare 10^3 to 2^3 than what you typed.

  5. I solved it a bit differently (less elegantly),, using a bit of algebra. If we say y = 10^(1/10) and x=2^(1/3), then that means x^3 = 2 and y^10 = 10, so if x^10 > 10, we know x is bigger.

    Since x^3 = 2, x^10 = x * x^9 = 8x.

    For x^10 > 10, then, we just need 8x > 10, or x > 10/8.

    So is x bigger or smaller than 10/8?

    Well, (10/8)^3 is 1000/512, which is clearly smaller than 2 (which is x^3). So x is indeed bigger than 10/8, and thus x > y.

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