I was given this question: suppose 123456789abcd is written in base 14. What is its remainder when divided by 13? I know the answer is zero but I had to do it using Wolframalpha. I'm wondering if there's a smart way to do it, instead of doing it by hand?
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5the same method as divisibility by $9$ when written in base $10,$ because $14 \equiv 1 \pmod{13}$ – Will Jagy Jul 18 '23 at 16:28
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What's this method? Can you provide a link? – user1691278 Jul 18 '23 at 16:30
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Sum its digits mod $14$ to get the remainder. – user3257842 Jul 18 '23 at 17:06
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As explained here in the dupe, this is the radix $14$ analog of casting out nines, by $14\equiv 1\pmod{13}.\ \ $ – Bill Dubuque Jul 19 '23 at 01:44
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The number $$a_n B^n + a_{n-1} B^{n-1} + ... + a_0 B^0$$ and the number $$a_n + a_{n-1} + ... + a_0$$ have the same remainder when dividing by $B-1$. The reason is that the difference between these two numbers is $$a_n(B^n - 1) + a_{n-1}(B^{n-1} - 1) + ... a_0( B^0 - 1)$$ and each $B^k - 1$ is divisible by $B - 1$ since $x - 1$ divides the polynomial $x^k - 1$.
Use $B = 14$ in the case at hand; the remainder when dividing by $13$ will be the same as the remainder of the sum of its digits in base $14$.
Zarrax
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Please strive not to post more (dupe) answers to dupes of FAQs (& PSQs), cf. recent site policy announcement here. – Bill Dubuque Jul 19 '23 at 01:45
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This is a case of omega divisibility: In base $n$, a number is divisible by $n-1$ if and only if the sum of its digits is. In our usual base-ten numbering system, this gives us the rule for divisibility by 9. And in base 14, it gives us a rule for divisibility by 13.
Adding up the digits in $123456789\text{abcd}_{14}$ gives (decimal) 91, which is a multiple of 13 ($13 \times 7$). Thus, the original number is a multiple of 13.
Dan
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Please strive not to post more (dupe) answers to dupes of FAQs (& PSQs), cf. recent site policy announcement here. – Bill Dubuque Jul 19 '23 at 01:45
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@BillDubuque: I would dispute that this question is a duplicate. Your linked "already has an answer" question is about divisibility by 9 in normal base-10 arithmetic, while this one is about divisibility by 13 in base-14. Sure, if you're familiar with alternate-base arithmetic, it's easy to recognize this as being basically the same problem, but this may not be obvious to beginners. – Dan Jul 19 '23 at 15:48
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The linked answer in the dupe closing comment (and also others, e.g. Arturo's) discuss the rules in arbitrary radix. It thwarts pedagogy and site organization to duplicate these simple arguments for random specific radices. – Bill Dubuque Jul 19 '23 at 16:29