The sum of ABCD and DCBA is always divisible by 11, where A, B, C and D are digits of a number. I understood that, ABCD = A(1000) + B (100) + C(10) + D(1) DCBA = D(1000) + C(100) + B(10) + A (1) Then ABCD + DBCA = A(1001) + B(110) + C (110) + D(1001) =11[91A + 10B +10C +91D] which is divisible by 11.

I want a generalized proof for the above problem. Also, I want why it is applicable only for 11 and why it is valid for even digit of numbers? Thanks in advance

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    $\begingroup$ If you ask a question that's related to another question you've asked (in this case math.stackexchange.com/questions/52361/…), it makes sense to link to the question so people get a better idea of the context; e.g. in this case Gortaur wouldn't have had to bother to tell you about the divisibility rule for 11. $\endgroup$ – joriki Jul 20 '11 at 10:14
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    $\begingroup$ For an odd number of digits the middle digit will lead to a summand 2 * d * 10^k, so it can only work for divisibility by multiples of 2 and/or 5. $\endgroup$ – starblue Jul 20 '11 at 10:47
  • $\begingroup$ In what direction do you want to generalize your result? $\endgroup$ – Américo Tavares Jul 21 '11 at 9:34

Do you know the following rule? Given a number $\overline{a_na_{n-1}...a_1}$, it is divisible by $11$ iff $a_n+a_{n-2}+... - a_{n-1}-a_{n-3}-... $ divisible by $11$. Say, $1001$ is divisible by $11$ since $1+0-0-1 = 0$ is divisible by $11$.

Now, if you write $\overline{b_{2n}b_{2n-1}...b_2b_1}$ and wonder about the divisibility of $b = \overline{b_{2n}b_{2n-1}...b_2b_1}+\overline{b_{1}b_{2}...b_{2n-1}b_{2n}}$ then you get $$ b = \sum\limits_{k=1}^{2n}b_k(10^k+10^{2n+1-k}) $$ and you only need to prove that $10^k+10^{2n+1-k}$ is divisible by $11$. I believe you can do it based on the criteria I gave above. Finally, to prove that it only hold for $2n$ you can conisder $10^2n$ which has $2n+1$ digit and $10^{2n}+1$ is not divisible by $11$.


I will prove a smaller bit, show how it matters, and state a more general idea.

Idea 1: A number consisting of the digit 1, an even number of zeroes, and ending in the digit 1 is divisible by 11.

Proof: 1000...01 = 1111...11 - 111...10 (e.g. 1001 = 1111 - 110). The former has an even number of (11)s, so it is clearly divisible by 11. The latter has one fewer (11), and is also divisible by 11 (and by 10, incidentally). This also answers your question: why an even number of digits? So that this is true.

Idea 2: When "symmetrically adding" a number with an even number of digits (e.g. ABCD + DCBA), the sum can be expressed as a sum of numbers consisting of a pair of one of the original digits (such as A, B, C, or D) separated by an even (possibly zero) number of zeroes, and possibly ending with a few more zeroes. So ABCD + DCBA = A00A + BB0 + CC0 + D00D.

This is proved naturally.

Putting these together yields the general proof: a number with an even number of digits that is "symmetrically added" to itself is always divisible by 11.

  • $\begingroup$ I note additionally that Idea 1 is often used to prove Gortaur's 'rule' without using modular arithmetic. I don't know why, but that's how I first came across it. $\endgroup$ – davidlowryduda Jul 20 '11 at 10:08

Corrected in response to Niel de Beaudrap's comment.

Main ideas:

  1. It is a consequence of the test for divisibility by $11$. The remaider of the division of a number $a$ by $11$ is equal to the remainder of the difference of the sum of the odd ordered digits of $a$ from the sum of the even ordered digits of $a$.
  2. And the base $10$ property: for some $n$ the power $10^n\equiv 1\pmod {11}$.
  3. $10\equiv (-(-10))\equiv -1\pmod{11}$


$$a=\overline{a_{n}a_{n-1}\cdots a_{2}a_{1}a_{0}}=a_{n}10^{n}+a_{n-1}10a^{n-1}+\cdots +a_{2}10^{2}+a_{1}10+a_{0}.$$

For $\mod 11$ we have $10^{0}\equiv 1,10^{1}\equiv 10,10^{2}\equiv 1$, and if $n$ even

$$10^{n}\equiv 1;$$

if $n$ odd $$10^{n}\equiv 10.$$ Thus

$$a\equiv (a_{0}+a_{2}+a_{4}+\cdots )+10(a_{1}+a_{3}+a_{5}+\cdots )\pmod{11}.$$

Since $-10\equiv 1\pmod{11}$, we have $$10\equiv (-(-10))\equiv -1\pmod{11}$$ and

$$a\equiv (a_{0}+a_{2}+a_{4}+\cdots )-(a_{1}+a_{3}+a_{5}+\cdots )\pmod{11}.$$

  • Your case is: For $a=A10^{3}+B10^{2}+C10+D$ $$a\equiv (D+B)-(C+A)\pmod{11}$$ and for $b=D10^{3}+C10^{2}+B10+A$ $$b\equiv (A+C)-(B+D)\pmod{11}$$ and $$a+b\equiv (D+B)-(C+A)+\left( (A+C)-(B+D)\right) \equiv 0\pmod{11}.$$

  • For $$a=A10^{4}+B10^{3}+C10^{2}+D10+E$$ and $$b=E10^{4}+D10^{3}+C10^{2}+B10+A,$$ we have $$a+b\equiv (A+C+E)-(B+D)+\left( (E+C+A)-(D+B)\right) \not\equiv 0\pmod{11}.$$

  • Can you generalize this?

Corrected in response to Niel de Beaudrap's comment.

I want a generalized proof for the above problem. I want why it is applicable only for 11 and why it is valid for even digit of numbers?

I am not sure in what direction does OP want to generalize his/her result. It is valid only for numbers with an even number of digits. (see above)

  • Here is a possible generalization of the proof above (radix $10$) for the radix $r$. Since

$$r^{0}\equiv 1\pmod {r+1},r^{1}\equiv r\pmod {r+1},$$

$$r^{2}\equiv 1\pmod{r+1},r^{3}\equiv r\pmod {r+1},\ldots$$

and $$ABCD_{r}\equiv (D+B)_{r}-(C+A)_{r}\pmod {r+1},$$

$$DCBA_{r}\equiv (A+C)_{r}-(B+D)_{r}\pmod {r+1},$$

we obtain

$$ABCD_{r}+DCBA_{r}\equiv 0\pmod {r+1}.$$

  • $\begingroup$ Sir,Thank you so much for your answer. If you don't mind, can I have your email ID? my email id: mahima113@rediffmail.comu $\endgroup$ – MahimA Aug 6 '11 at 8:39
  • $\begingroup$ @MahimA: You are welcome! You can see it on my blog $\endgroup$ – Américo Tavares Aug 6 '11 at 8:49
  • $\begingroup$ It is not valid for all numbers with an odd number of digits. Counterexample: 123 + 321 = 444, whose residue modulo 11 is 4. $\endgroup$ – Niel de Beaudrap Aug 15 '11 at 16:06
  • $\begingroup$ @Niel de Beaudrap You are right. Thank you very much! I corrected the post. $\endgroup$ – Américo Tavares Aug 15 '11 at 16:42
  • $\begingroup$ @MahimA: As commented by Niel de Beaudrap the result is not valid for the case when the number of digits is odd. I corrected post. $\endgroup$ – Américo Tavares Aug 15 '11 at 16:50

This follows quite straightforwardly from my answer to your prior question. Recall that the radix $\rm\:b\:$ digit string $\rm\ d_n\ \cdots\ d_1\ d_0\ $ denotes a polynomial expression $\rm\ P(b)\ =\ d_n\ b^n +\:\cdots\: + d_1\ b + d_0\:,\ $ where $\rm\:\ P(x)\: =\ d_n\ x^n +\:\cdots\: d_1\ x + d_0\:.\: $ The reversed digits polynomial is $\rm\ {\bar P}(x) = x^n\ P(1/x)\:.\:$ So if $\rm\:n\:$ is odd then $\rm\: mod\ \ b+1:\ \ b\equiv -1\ \Rightarrow\ {\bar P}(b) = b^n\ P(1/b) \equiv (-1)^n P(-1)\equiv {-}P(-1)\:,\:$ therefore $\rm\: P(b) + \bar P(b)\equiv P(-1)-P(-1)\equiv 0\:,\: $ as claimed.

REMARK $\ $ Notice how simple it becomes once one recognizes the innate polynomial structure. Whenever something has polynomial form, this structure should be brought to the fore so that ubiquitous knowledge of polynomials may be exploited to the hilt.


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