Is an arbitrary number of the form xyzxyz divisible by 7, 11, 13?
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
$\endgroup$ 18 Answers
$\begingroup$Hint:
$$7\cdot11\cdot13=1001$$
$\endgroup$ 6 $\begingroup$Every number of that form is divisible by $7$, $11$ and $13$:
$$\underbrace{xyz}_\text{1000xyz}~xyz = 1000xyz + xyz = 1001xyz = 7\cdot11\cdot13\cdot xyz$$
$\endgroup$ $\begingroup$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 $13$.
$\endgroup$ 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.
$\endgroup$ $\begingroup$Hint:
Any number of the form of $xyzxyz$ is some integral multiple of $1001=7\cdot 11\cdot 13$
QIf the 6-digit number 479xyz is exactly divisible by 7, 11 and 13, then
Problem Solving : Any number is divisible by 7, 11, 13 when it is written in the form of xyzxyz. Hence x =4, y=7, z=9. We can take any whole number for value of x,y and z
Why & How ?
Explanation step-by -step
First, Let's find the LCM of 7,11,13, i.e. 1001
That 6-digit number must be a multiple of 1001.
Now assume x= 0, y=0 and z=0.
Then 479xyz will be 479000. Next divide 479000 by 1001 and get quotient 478 and remainder 522. Now if we subtract 522 from 479000 the result is 478478 ( Divisible by 1001, or 7, 11 or 13 ). But here starting 3 digits are not 479.
So, we require the next multiple of 1001 i.e. 479 Hence , the number must be 1001 x 479 = 479479, this number satisfies all of the given conditions.
Therefore x =4,y=7, z=9
Hence, We can take any 1-digit whole number for value of x,y and z
$\endgroup$ $\begingroup$\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 $\overline{xyzxyz} = 7\cdot11\cdot13\cdot xyz$
$\endgroup$ 1 $\begingroup$Here's a handy proof to remember for divisibility by 7,11 and 13.
Remember lcm(7,11,13) = 7 * 11 * 13 since all the three numbers are co-primes.
Any integer n can be written as n = 1000q+r ---{ 0=<r<=999]
So, n = 1000(q - r) + 1001r
Since 7 * 11 * 13 = 1001.
7 * 11 * 13 | q-r -- [ if 7 * 11 * 13 | n then 7 * 11 * 13 | 1000(q-r) and 7 * 11 * 13 doesn't divide 1000.]
Hence for any integer to be divisible by 7 * 11 *13 q -r should be divisible by the product, where q and r are the quotients and remainders of the number n when divided by 1000.
Now, just divide xyzxyz by 1000 and notice that both the remainders and quotients are xyz, respectively.
Hence, 7, 11 and 13 | xyzxyz.
