Browsing category ninja maths

The Mathematical Ninja and an Irrational Power

“The square root of two… I don’t even know how to say this. The square root of two to the square root of threeth power?” “$\sqrt{2}^{\sqrt{3}}$?” said the Mathematical Ninja. “I wouldn’t bother saying it, I’d just write it down.” “But what does it mean? I mean, I can just

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“How many days old are you?”

“How old are you?” asked young Fred. (This is not, technically, an ‘Ask Uncle Colin’. It’s an ‘Ask Daddy’.) “I’m 42, buddy.” “42 days?” “No, 42 years!” “Oh. But how many days is that?” It is not quite 7:15am on New Year’s Day. I have not yet had my coffee.

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The Mathematical Ninja and the Cube Root of 13

A physicist. A calculator. The Mathematical Ninja’s face - what could be seen of it - was more snarl than feature. It’s quite tricky to hiss something that doesn’t have any sibilant consonants, but they hissed all the same: “The cube root of 13? You don’t need a calculator for

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The Mathematical Ninja and the *Other* Pole

“Sensei, why have you covered the entire Earth in an area-preserving wrap?” “It’s all @colinthemathmo's doing.” “I’m surprised you’re doing it in hardware rather than working it out in your head.” “Oh, $\frac{1000}{\sqrt{\pi}}$? That’s trivial.” “But of course it is.” “I mean, $\frac{1}{\pi}$ is pretty close to $\frac{1}{\sqrt{10}}$, which is

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Calculating $e^e$ and $e^{-\frac{1}{e}}$

"The Mathematical Ninja is currently on sabbatical. Leave a message after the tone... or else!" Oh dear! How are we going to figure out $e^e$ now? Let alone $e^{-\frac{1}{e}}$? We'll just have to roll up our sleeves and get our thinking hats on, that's all. OK, $e^e$ First of all,

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The Mathematical Ninja and the Unknown Powers

The Mathematical Ninja peered at the problem sheet:   Given that $(1+ax)^n = 1 - 12x + 63x^2 + \dots$, find the values of a and n   Barked: “$n=-8$ and $a=\frac{3}{2}$.” The student sighed. “I get no marks if I just write down the answer.” Snarled: “You get no

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A logs puzzle

Via @markritchings, an excellent logs problem: If $a = \log_{14}(7)$ and $b = \log_{14}(5)$, find $\log_{35}(28)$ in terms of $a$ and $b$. One of the reasons I like this puzzle is that I did it a somewhat brutal way, and once I had the answer, a much neater way jumped

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A Challenge to the Mathematical Ninja

“I beg your pardon?!” yelled the Mathematical Ninja. The terribly well-dressed gentleman stood his ground. “I said, sensei, I would work $0.8^{10}$ out differently.” A sarcastic laugh. “This, I have to see!” “Well, $8^{10} = 2^{30}$, which is about $10^{9}$.” “About.” “Obviously, we can do better with the binomial: $2^{10}$

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The Mathematical Ninja and $\arctan(0.4)$

It took the Mathematical Ninja a little longer than normal; the student had managed to rummage around in her bag and lay a finger on the calculator before simultaneously feeling her arm pulled away by a lasso and hearing "0.3805. Or, as a one-off, since the question is asking for

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Powers

“Here’s a quick one,” suggested a fellow tutor. “Prove that $2^{50} < 3^{33}$.” Easy, I thought: but I knew better than to say it aloud. First approach “I know that $9 > 8$,” I said, checking on my fingers. “So if $2^3 < 3^2$, then $2^{150} < 3^{100}$ and $2^{50}

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