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Divisibility rule for 22: Under what conditions a natural number $N$ is divisible by $22$ ?

My thought is

The divisibility rule for $22$ is that the number is divisible by $2$ and by $11$. Divisibility by $2$ requires that the number ends in $0$, $2$, $4$, $6$ or $8$. Divisibility by $11$ requires that the difference between the sum of the the digits in odd positions and the sum of all the digits in even positions is $0$ or divisible by $11$.

how can i prove that ?

my attempt:

Indeed,

$2 \mid N$ then $N = 2.k$ , and $11\mid N\ $ i.e. $11 \mid 2.k$

or $\textrm{gcd}(2,11) = 1$ then by Lemma gauss $11\mid k$, i.e. $k=11p$ where $N = 2k = 2.11. p = 22p$

Thus

$22\mid N$

Is my reasonable right ?

Educ
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    is $b=1$? I asked that because you have $N=a_{n}a_{n-1}\cdots a_{2}a_{1}a_{0}$ and $N=a_{n}b^{n}+a_{n-1}ba^{n-1}+\cdots +a_{2}b^{2}+a_{1}b+a_{0}$ – Jlamprong Sep 03 '14 at 07:29
  • If you understand where the individual rules regarding the divisibility of $2$ and $11$ come from, you can just combine them by observing that if $2|N$ and $11|N$ then $22|N$. – Kim Jong Un Sep 03 '14 at 07:29
  • Related : http://math.stackexchange.com/questions/328562/divisibility-criteria-for-7-11-13-17-19 – lab bhattacharjee Sep 03 '14 at 07:51
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    @labbhattacharjee Thanks but i'm not newbie in http://math.stackexchange.com/ i know how i search in it to find related subject – Educ Sep 03 '14 at 08:07
  • You have already asked questions about divisibility by 11. The answer for divisibility by 2 is obvious. So yes, the answer you give for 22 is correct. But your proof is a mess. The proof is simple: a number is divisible by 22 if and only if it is divisible by 2 and by 11. – almagest Sep 03 '14 at 09:11
  • see i update my attempt – Educ Sep 03 '14 at 12:33
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    The rule for divisibility by 22 is, you divide by 22. If the remainder is zero, the number is divisible by 22; if not, then not. – Gerry Myerson Sep 03 '14 at 12:55
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    Simpler: $,11k,$ even $,\Rightarrow,k,$ even, so $,2,11\mid n,\Rightarrow,2\cdot 11\mid n.\ $ Generally $$,a,b\mid c\iff {\rm lcm}(a,b)\mid c,$$ the universal property/definition of $,\rm lcm$. – Bill Dubuque Sep 03 '14 at 14:27
  • @BillDubuque Thanks but what about gcd am i right ? – Educ Sep 03 '14 at 14:56
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    @Educ $\ $ Generally $\ \gcd(a,b) = 1,\Rightarrow,{\rm lcm}(a,b) = ab.\ $ In fact $\ \gcd(a,b),{\rm lcm}(a,b) = ab.\ \ $ – Bill Dubuque Sep 03 '14 at 15:01
  • @BillDubuque yes i see Thanks so much – Educ Sep 03 '14 at 15:05
  • @BillDubuque you can write what you said as answer and 'ill accept because it's very simpler way than mine – Educ Sep 03 '14 at 15:07
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    @Educ Ok, I added an answer. – Bill Dubuque Sep 03 '14 at 15:21

2 Answers2

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Yes, that's correct. After applying the divisibility tests for $\,2,11\,$ we know that $\,2,11\mid n,\,$ so it follows that $\,2\cdot 11\mid n,\,$ by Euclid's lemma, or uniqueness of prime factorizations, etc. But one can deduce this much more simply by parity. Namely $\,11\mid n\,\Rightarrow\, n = 11k.\,$ Since further $\,n = 11k\,$ is even, we deduce by parity arithmetic that $\,k\,$ is even $\, k = 2j,\,$ so $\,n = 11k = 11(2j),\ $ thus $\,2\cdot 11\mid n.$

Generally: $\,\ a,b\mid n\iff {\rm lcm}(a,b)\mid n,\ $ the universal property/definition of $\,\rm lcm.$

And $\ \gcd(a,b) = 1\iff {\rm lcm}(a,b) = ab,\ $ which is a special case of $\ \gcd(a,b)\,{\rm lcm}(a,b) = ab.$

Bill Dubuque
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The divisibility rule for 22 is that the number is divisible by 2 and by 11. Divisibility by 2 requires that the number ends in 0, 2, 4, 6 or 8. Divisibility by 11 requires that the difference between the sum of the the digits in odd positions and the sum of all the digits in even positions is 0 or divisible by 11.

Lakshita Kaushik
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