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It is asked:

What is the remainder when $25$ divides a number where natural ones are arranged in order from $1$ to $35$ as a single number $1234567...333435$?

Before showing the way I did it, let's consider this: since $12=6\cdot 2$ it divides $9216$, coming from divisibility tests for $6$ and $2$ which the number $9216$ satisfies.

As I remember, when evaluating if $25$ divides $123...35$, I didn't use this method (firstly), instead, just picked the last 2 digits, $35$ and divided it into $25$: $35=1\cdot 25+10.$ So I concluded that when $123...35$ is divided by $25$, it will also have the remainder $10$, which is correct. Another neat way to show it is by writing $123..35=123...333425+10$. Here, the number $123...25$ divides into $25$ evenly, since the last 2 digits, $25$ also divides into $25$ evenly, so $10$ above indicates the remainder.

And here is the origin of confusion: $25=5\cdot 5$ so, to see if $25$ divides $123...35$ evenly, the latter huge number ($123...35$) must divide into $5$ evenly, which it does, and one can conclude that $123...35$ divided by $25$ won't have any remainder, which is untrue. By the way, it implies as well, that the case with $12$ further above in the first paragraph is not always applicable in other cases (not universal).

But I also don't trust my method completely in that it is universal, consider this apparent proof: when evaluating if $16$ divides $114664$ evenly, picking the last digits, $64$ and dividing it by $16$ or by $4,$ (because $16=4^2$) doesn't produce any remainder. But concluding $114664$ divided by $16$ either doesn't have a remainder is erroneous.

Finally, my questions are:

1) what is your thoughts on my way of solving the problem?

2) what is the correct (and universal) way of solving the problem?

3) is not using the method which is used in evaluating if $12$ divides $9216$ evenly when solving like problems saying evaluate if $x$ divides into $y$ evenly universal? (Though I believe it is not universal, I want to hear from you.)

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    $\begingroup$ $12$ does not divide $13626$. The remainder on division is $6$. $\endgroup$ – Eric Towers Sep 9 '17 at 19:17
  • $\begingroup$ @EricTowers Thanks, changed it to 9216. $\endgroup$ – user36339 Sep 9 '17 at 19:20
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    $\begingroup$ Do you realize that this failure fully answered your question? $\endgroup$ – Eric Towers Sep 9 '17 at 19:20
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1) As you have edited away, both $2$ and $6$ divide $13626$, but $12$ does not. All you can conclude from divisibility by both $2$ and $6$ is divisibility by the least common multiple of $2$ and $6$, which is $6$.

2) A way to follow a version of your idea is this: $12 = 6 \cdot 2$ and $6$ divides $13626$, so we must ask: does $2$ divide $13626/6 = 2271$. It does not, so $12$ does not divide $13626$.

With your replacing example, $6$ divides $9216$, so we must ask: does $2$ divide $9216/6 = 1536$. It does, so $12$ does divide $9216$.

In the context of the original problem, $25 = 5 \cdot 5$. We see that $123\cdots 3435$ is divisible by $5$, so we see whether $246913\cdots 66687$ is divisible by $5$, which fails. So, $123\cdots 3435$ is not divisible by $25$.

3) No. As explained above, for $x, y, n$ integers with $x$ dividing $n$ and $y$ dividing $n$, all we can conclude is that the least common multiple of $x$ and $y$ divides $n$.

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since $12=6\cdot 2$ it divides $13626$, coming from divisibility tests for $6$ and $2$ which the number $13626$ satisfies.

Unsound (as pointed out in the comments) and invalid argument. One would conclude that $12$ divides $18$ from that.

You can only split one divisibility test into two if the two numbers are co-prime.


And here is the origin of confusion: $25=5\cdot 5,$ so, to see if $25$ divides $123...35$ evenly, the latter huge number ($123...35$) must divide into $5$ evenly, which it does, and one can conclude that $123...35$ divided by $25$ won't have any remainder, which is untrue.

No, one cannot conclue that, per the reason stated in the above section.


1) what is your thoughts on my way of solving the problem?

Understandable but nonsensical.

2) what is the correct (and universal) way of solving the problem?

As you have said, noting that the huge number minus $10$ is divisible by $25$.

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  • $\begingroup$ Thanks! I fixed my error, seems like I hurried a bit in asking the question. $\endgroup$ – user36339 Sep 9 '17 at 19:22
  • $\begingroup$ @Tug'tekin The argument is still invalid. $\endgroup$ – Kenny Lau Sep 9 '17 at 19:23
  • $\begingroup$ $9216=768\cdot12$ I know that you know it, but I wrote it because you said "unsound" $\endgroup$ – user36339 Sep 9 '17 at 19:26
  • $\begingroup$ @Tug'tekin doesn't make the argument valid. $\endgroup$ – Kenny Lau Sep 9 '17 at 19:26
  • $\begingroup$ Sound vs valid (sorry, even I got confused and interchanged both terms) $\endgroup$ – Kenny Lau Sep 9 '17 at 19:27
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since 12=6⋅2 it divides 9216, coming from divisibility tests for 6 and 2 which the number 9216 satisfies. since 12=6⋅2 it divides 9216, coming from divisibility tests for 6 and 2 which the number 9216 satisfies.

Your this technique is totally wrong , $$$$ a number 'x' is said divisible by z and y $$$$ after using their divisibility tests individually $$$$ only and only if $$$$ z and y are coprimes. $$$$ And 6 and 2 are not coprimes. $$$$ You can write $12= 4×3$ (4 &3 are coprimes) $$$$ and then do individual tests on 9216 $$$$ it satisfies both, therefore 9216 is divisible by 12.

THE CORRECT WAY

We can note the given no. as $$\sum_{x=35}^{10} 10^{2(35-x)} x + \sum_{x=9}^{1} 10^{61-x} x$$ We can also write it as $$35 +100× 34 +10000 A$$ Note that A's last digit is 3, $$5 (7+ 680 +1000(2A))$$ $$ 5 (687+1000 (2A))$$ $$5 (12+ 5 C)$$ $$60 + 5C$$ $$ 10 + 5D$$

Thus the given no. is not divisible by 25 and lives remainder of 10 behind.

I didn't understand what you did, your description was kind of confusing for me. So i don't know if your technique is right or not, i will update soon.

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  • $\begingroup$ The number you have written as a sum is $113\,580\,246\,902\,345\,679\,012\,345\,675$, which is less that the square root of the OP's number. The first sum is $112345679012345679012345675$, which is (I suspect) very much not what you were thinking it would be. $\endgroup$ – Eric Towers Sep 9 '17 at 19:32
  • $\begingroup$ Your edit no longer gives an integer. Note that $10^{35 - 2x}$ is less than $1$ for $x\geq 18$. $\endgroup$ – Eric Towers Sep 9 '17 at 19:36
  • $\begingroup$ Just corrected that $\endgroup$ – neonpokharkar Sep 9 '17 at 19:39
  • $\begingroup$ Thanks by the way $\endgroup$ – neonpokharkar Sep 9 '17 at 19:40
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    $\begingroup$ "2(35-x)" and "61-x" actually works. $\endgroup$ – Eric Towers Sep 9 '17 at 19:45

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