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Pardon my English as I'm not a native speaker of the language and I'm not a big math guy, so please bear with me and my ignorance for a bit. I've unconsciously stumbled upon something that's most probably blatantly obvious and easy, yet I have no idea how to explain it. So here it goes:

$$11! = 39916800$$ $$10! = 3628800$$ $$11! - 10! = 39916800 - 3628800 = 36288000$$

As you can see, the result equals the value that $10!$ returns "plus" an extra digit added. Is this just a cool coincidence or is there any logic behind that I'm just not aware of?

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migrated from mathematica.stackexchange.com Aug 11 '18 at 4:25

This question came from our site for users of Wolfram Mathematica.

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$11!-10!=11\times 10!-1\times10!=10\times10!$

"10 times an integer" means that we put an extra 0 after the last digit of the integer.

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    $\begingroup$ Just a note to OP: This approach is also applicable to $(10n+1)!-(10n)!$ for any positive integer $n$, but an extra factor of $n$ will be applied to the number. $\endgroup$ – Mythomorphic Aug 11 '18 at 4:49
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By definition of the factorial, we have $$11! - 10! = 11 \cdot 10! - 10! = (11 - 1) 10! = 10 \cdot 10!$$ Now, as you may remember, multiplying by $10$ (or whatever base you are in) has the effect of shifting all digits to the left and adding a zero at the right of your number. This perfectly explains your extra zero at the end.

You might try to come up with other examples, such as $101! - 100!$.

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