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I know that 26=13.2. The last digit of the number is 4, it is divisible by 2. I have to remove the last digit and check out if the remaining digits are divisible by 13 or?

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  • $\begingroup$ what do you mean by '2 digits in front of the number 14'? $\endgroup$ – oliverjones Apr 20 '17 at 18:55
  • $\begingroup$ Do you mean $14xx$ or $xx14$ or something different? $\endgroup$ – Mark Bennet Apr 20 '17 at 18:56
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    $\begingroup$ gotta be xx14 -- two digits in front of the number 14 $\endgroup$ – gt6989b Apr 20 '17 at 18:56
  • $\begingroup$ I wanted to say xx14. $\endgroup$ – Elena Apr 20 '17 at 18:58
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You are being asked to find an integer $x$ such that $10 \leq x \leq 99$ and $$ 100 x + 14 \equiv 0 \pmod{26}. $$ Solve this equation, and $x$ will be the two digits you want. There are multiple possible answers.


A good first step is to rewrite the equation in a more convenient form. Since it is modulo $26,$ you can replace any of the numbers by something equivalent modulo $26.$ For example, $100 \equiv 22 \pmod{26},$ so you can write $22$ instead of $100.$ I think it's more convenient to use the fact that $100 \equiv -4 \pmod{26},$ however; it lets you work with smaller numbers.

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  • $\begingroup$ i think the OP is asking how to solve this equation $\endgroup$ – gt6989b Apr 20 '17 at 19:00
  • $\begingroup$ We know that $100x + 14 \equiv 0\pmod{26} \iff 100x + 14 = 26m$, for some integer $m$. Thus, $$x = {26m-14\over 100} = {13m - 7 \over 50}.$$ Since we require that $10\le x \le 99$, choosing $m$ appropriately should give us the desired result. For example, we can choose $m=39$. Then we have $500/50=10$, which implies $x=10$ would be a good starting point, and so $1014$ will be a satisfactory answer. $\endgroup$ – Decaf-Math Apr 20 '17 at 19:05
  • $\begingroup$ @gt6989b I think that looking at the question this way is fundamentally different from what the OP was trying to do: "remove the last digit and check if the remaining digits are divisible by ..." etc. But I added some detail in hope that it may help. $\endgroup$ – David K Apr 20 '17 at 19:06
  • $\begingroup$ How many answers do I have to get? 2? The first one is 1014 $\endgroup$ – Elena Apr 20 '17 at 22:03
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    $\begingroup$ I don't know if you need more than one answer, but you have a choice. Since 1300 is the smallest multiple of 100 divisible by 26, you can add any positive multiple of 1300 to 1014 and get another solution, unless the multiple is so large that the result is greater than 9999. It turns out there is room for seven separate solutions between 1000 and 9999, starting with 1014 and ending with 8814. $\endgroup$ – David K Apr 20 '17 at 22:21
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Starting with $100 x + 14 \equiv 0 \pmod{26}$ you can take $100 \pmod{26}$ to get $74, 48, 22, -4$, etc. $-4$ is the smallest, so let's try that.

Now the equivalence to solve is $-4x + 14 \equiv 0 \pmod{26}$. It'll be easier to see the solution if the LHS is positive, so add $26$ to get $-4x + 40 \equiv 0 \pmod{26}$. Is it possible to subtract 4 from 40 a few times to get $26$? If you subtract $4*3=12$ you still have $2$ to go to get to $26$.

So add another $26$ to the equation to get $$-4x + 40 \equiv 0 \pmod{26}$$. Note that $66-26=40=4*10$. So $x$ is $10$.

You would then verify $x=10$ in the original problem. Is $1014 \equiv 0 \pmod{26}$?

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  • $\begingroup$ Yes, it is. But I think a have more than 1 answer because (100, 26)=2 $\endgroup$ – Elena Apr 20 '17 at 22:09
  • $\begingroup$ I don't understand your comment. There will be $7$ answers. The answer can be any value of the form $10+13x$, with $0 \leq x \leq 6$. I interpreted the problem as requiring just one answer. $\endgroup$ – Χpẘ Apr 20 '17 at 22:24
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    $\begingroup$ I guess not. David K also specifies $7$ answers in the comments to his answer. $\endgroup$ – Χpẘ Apr 20 '17 at 22:29
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Let $1\le a\le 9$ and $0\le b\le9$ be those two front numbers ($ab14$). So, our desired number is $10^3a+10^2b+14$ where $$10^3a+10^2b+14\equiv0\pmod{26}\implies 12a-4b\equiv12\pmod{26}\implies 3a-3=3(a-1)\equiv b\pmod{13}$$ All choices that satisfy this (just plug in $a=1,2,\ldots9$ and see what happens):

$1014$ $2314$ $3614$ $4914$ $6214$ $7514$ $8814$

Notice how in the congruence, I used $$a\equiv b\pmod{m}\implies \frac{a}{c}\equiv \frac{b}{c}\pmod{\frac{m}{\gcd(m,c)}}$$ as long as $c\vert a$ and $c\vert b$. Also notice that $a\equiv9$ is not possible because that gives an invalid number for $b$

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  • $\begingroup$ I thought we only had 1 answer - 1014? Or? $\endgroup$ – Elena Apr 20 '17 at 19:49
  • $\begingroup$ You can check for yourself that all $7$ of those integers work just fine. $\endgroup$ – user12345 Apr 20 '17 at 22:51
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One way to find a solution is to start from the useful and memorable fact that $7 \times 11 \times 13 = 1001$. Hence $26$ divides any even multiple of $1001$, in particular $4004$.

If we can find $x10$ divisible by $26$, then $4004 + x10$ will yield a solution. $x10$ is divisible (for any $x$) by $2$, so it suffices to find $x1$ divisible by $13$. Since $91 = 7 \times 13$, a solution is $4004 + 910 = 4914$.

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