13
$\begingroup$

Why does it seems like every number $ababab$, where $a$ and $b$ are integers $[0, 9]$ is divisible by $13$?

Ex: $747474$, $101010$, $777777$, $989898$, etc...

$\endgroup$

9 Answers 9

20
$\begingroup$

Note that $$ [ababab] = a\times 101010 + b \times 10101 = 13 (7770a + 777b) $$


Also noteworthy: $$ 10101 = 1 + 10^2 + 10^4 \equiv \\ 1 + 3^2 + 3^4 = 1 + 9 + 9^2 \equiv\\ 1 +(-4) + (-4)^2 = 1 - 4 + 16 = 13 \equiv 0 $$ where $\equiv$ indicates equivalence modulo $13$.

$\endgroup$
4
  • $\begingroup$ Can you provide an intuition on how you got to that decomposition? $\endgroup$
    – Juan
    Commented Jun 7, 2016 at 20:57
  • $\begingroup$ @Juan you can look at my answer for the intuition or simply multiply and adding everything in the second line of this answer will tell you the intuition. $\endgroup$ Commented Jun 8, 2016 at 2:23
  • $\begingroup$ @Juan I explain how it arises in my answer. $\endgroup$ Commented Jun 20, 2016 at 19:53
  • $\begingroup$ I guess you want $3^2=9$, not $3^2=9^2.$ $\endgroup$
    – user940
    Commented Jun 20, 2016 at 19:55
14
$\begingroup$

These numbers are of the form $(10a+b)\cdot 10101$, and $10101=13\cdot 777$.

$\endgroup$
2
  • $\begingroup$ So they are also all divisible by 7 and 37! $\endgroup$ Commented Jun 7, 2016 at 15:52
  • $\begingroup$ @BenBlum-Smith Indeed, and by 3. $\endgroup$ Commented Jun 7, 2016 at 15:55
5
$\begingroup$

Note: $$ab=a\times 10+b\times 1$$

so $$ab\times 100=ab00$$ therefore

$$abab=ab00+ab=ab(100+1)=ab\times 101$$ $$abab00=ab\times101\times100=ab\times10100$$ $$ababab=ab\times10101=ab\times3\times7\times13\times37$$

Hence it is divisible by $3$,$7,13$ and $37$.

$\endgroup$
4
$\begingroup$

A number $ABCDEF$ is divisible by $13$ if and only if $ABC-DEF$ is divisible by $13$.

Note that $\small{ABA-BAB=100(A-B)+10(B-A)+1(A-B)=91A-91B=13(7A-7B)}$.

Therefore $ABA-BAB$ is divisible by $13$.

Therefore $ABABAB$ is divisible by $13$.

$\endgroup$
1
  • 1
    $\begingroup$ "ABC−DEF is divisible by 13". (Hint of) proof? Or link to? $\endgroup$ Commented Jun 7, 2016 at 19:48
3
$\begingroup$

And not only that: Each such number is also divisible by the other primes 3, 7 and 37. Just factor the number 10101! Your specimen number is any 2-digit number $\times 10101\,$.

$\endgroup$
3
$\begingroup$

$abcabc=abc(10^3+10^6)=abc\cdot 1001000=abc\cdot13\cdot77000$.

REMARK.-It is easy generalisable to $abcabcabcabcabc.....abcabc$ for $2n$ times abc.

$\endgroup$
2
$\begingroup$

$ababab$ is a multiple of $10101$, which in turn is a multiple of $13$.

$\endgroup$
2
$\begingroup$

Hint $\,\ {\rm mod}\ 13\!:\,\ 10\equiv 6^2\,\Rightarrow\, 10^6\equiv 6^{12}\equiv 1\,$ by little Fermat.

Hence $\, 0\equiv 10^6-1 \equiv (10^2\!-1)(10^4\!+10^2\!+1) \equiv 99\cdot 10101$

Thus $\,99\not\equiv 0\,\Rightarrow\,10101\equiv 0\,\Rightarrow\, ab\cdot 10101 = ababab\equiv 0,\,$ for $\, 0 \le ab \le 99$

$\endgroup$
1
$\begingroup$

$100^0 \equiv 1 \bmod 13$

$100^1 \equiv 9 \bmod 13$

$100^2 \equiv 3 \bmod 13$

$(ababab)_{10}=c100^2+c100+c =c(100^2+100+1) \equiv c(3+9+1) \equiv 0 \bmod 13$, where $c=10a+b$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .