17
$\begingroup$

So I was given this question

Choose any 3-digit number xyz and write it after itself as follows: xyzxyz. Check whether it is divisible by 7,11, 13. Is an arbitrary number of the form xyzxyz divisible by 7, 11, 13?

I am completely lost by this question. I seen divisibility of prime numbers and how to work with it, but I'm unsure how to apply it to this problem

$\endgroup$
  • $\begingroup$ There're simple criteria of whole numbers divisibility by 2,3 and 5. The fact in OP can be used to check divisibility by the next prime numbers $7,11$ and $13$ (in decimal base representation of numbers), (see Hardy's number theory book). There's a sketchy video on YouTube on how to divide by $19$... $\endgroup$ – DVD Dec 21 '15 at 21:41
59
$\begingroup$

Hint:

$$7\cdot11\cdot13=1001$$

$\endgroup$
  • 4
    $\begingroup$ so 6 digit number = xyzxyz = xyz * 1001 = xyz * 7 * 11* 13 as 7*11*13 = 1001 other factors are 7 * 11 = 77, 11*13 = 143 , 7 * 13= 91 and factors of xyz. Would this be correct? $\endgroup$ – Micky Oct 28 '15 at 15:20
  • 1
    $\begingroup$ Exactly. Also, products of factors of $xyz$ and 7,11,13 respectively are the remaining factors. $\endgroup$ – Mythomorphic Oct 28 '15 at 15:23
  • 6
    $\begingroup$ That's certainly a good and useful trick/fact to memorize. $\endgroup$ – Mythomorphic Oct 28 '15 at 15:26
  • $\begingroup$ Oh wow, I love this hint - got my head tingling :) $\endgroup$ – nullpotent Oct 29 '15 at 16:11
  • 1
    $\begingroup$ This is one of two useful tricks that I was once taught. The other one is 27 * 37 = 999. $\endgroup$ – Jos Oct 30 '15 at 12:20
22
$\begingroup$

Every number of that form is divisible by $7$, $11$ and $13$:

$$\underbrace{xyz}_\text{1000xyz}~xyz = 1000xyz + xyz = 1001xyz = 7\cdot11\cdot13\cdot xyz$$

$\endgroup$
6
$\begingroup$

Here is a more naive approach. Just using the usual criteria.

Divisibility by 7: $$z+3y+2x+6z+4y+5x=7z+7y+7x=7 (x+y+z), $$so $xyzxyz $ is a multiple of $7$.

Divisibility by $11$: $$z-y+x-z+y-x=0, $$ so $xyzxyz $ is a multiple of $11$.

Divisibility by $13$: $$z-3y-4x-z+3y+4x=0, $$ so $xyzxyz $ is a multiple of $13$.

$\endgroup$
  • $\begingroup$ Wow. I would never have considered to do it this way. Though this way is not practical enough i still upvote just for your approach $\endgroup$ – Shailesh Oct 29 '15 at 13:17
  • $\begingroup$ If you're going that route, then the "blocks of 3" rule for 7 and 13 is even easier. xyz-xyz=0 $\endgroup$ – MattPutnam Oct 29 '15 at 15:53
  • $\begingroup$ Good point. I had not thought about these rules for many years so I only remembered the most basic ones (i.e., just replacing each power of 10 by its remainder). $\endgroup$ – Martin Argerami Oct 29 '15 at 16:52
  • 1
    $\begingroup$ @MattPutnam I guess the "blocks of three" rule only comes from what this task is supposed to show. $\endgroup$ – Paŭlo Ebermann Oct 29 '15 at 23:06
4
$\begingroup$

Existing answers are correct. This is an attempt to add a way of finding the answer that might help with other problems.

The first thing I asked myself was "What divisibility-related properties do numbers of the form xyzxyz have in common?". The obvious answer is divisibility by 1001.

I then looked at 7, 11, and 13.

Because $$7\cdot1\cdot3$$ is 21, their product must end with "1".

The product of three numbers, one slightly less than 10 and the other two slightly greater than 10 is in the general neighborhood of 1000.

That led to the easily checked conjecture that their product is 1001.

$\endgroup$
2
$\begingroup$

Hint:
Any number of the form of $xyzxyz$ is some integral multiple of $1001=7\cdot 11\cdot 13$

$\endgroup$
0
$\begingroup$

\begin{align} xyzxyz &= 10^5x + 10^4y + 10^3z + 10^2x + 10y + z\\ &= (10^5+10^2)x + (10^4+10)y + (10^3+1)z\\ &= 100(1001)x + 10(1001)y + (1001)z\\ &= 1001 (100x + 10y + z) \end{align}

Also $$\overline{xyz000} + xyz = \overline{xyxxyz} = 1001 * xyz$$

And $1001 = 7.11.13$

So any number of form $\overline{xyzxyz} = 7\cdot11\cdot13\cdot xyz$

$\endgroup$
  • $\begingroup$ @hkmather802 I assume you've used $\overline{ab}$ to mean $10a+b$, but in which case all $xyz$ should really be $\overline{xyz}$. $\endgroup$ – Mark Hurd Nov 3 '15 at 0:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.