# whole numbers and division

Consider the whole number with one thousand digits that can be formed by writing the digits 2772 two hundred and fifty time in succession. Is it divisible by 9? Is it divisible by 11?

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I answered yes to both because 2772 is divisible by 9 and 11. I am just trying to make sure this is correct! –  SNS Feb 27 '12 at 23:19
If the sum of the digits of $n$ is divisible by 9, then (and only then) $n$ is divisible by 9. From this, your number is divisible by 9 (the sum of its digits is $250\cdot18$). I've forgotten the divisibility test for 11... –  David Mitra Feb 27 '12 at 23:21
Can you show that your number is divisible by 2772? And can you finish up from there? –  Gerry Myerson Feb 27 '12 at 23:26
2772/9=308 2772/11=252 I think that if the original number is divisible by 9 and 11 then if you continue adding the same numbers over and over it should always be divisible by 9 and 11. This is the part I want to double check on. –  SNS Feb 27 '12 at 23:29
@DavidMitra If the alternating sum of the digits (add, subtract, add, subtract, etc.) is divisible by 11 then the number is divisible by 11. This is because $10^n\equiv (-1)^n\mod 10$. –  Alex Becker Feb 27 '12 at 23:29

The answer is yes; but, in my opinion, you did not give enough information in your comment for a justification.

One way to show it is to use the divisibility tests for 9 and 11.

Let's call your number, obtained by writing "$2772$" two hundred and fifty times in succession, $y$.

A number $n$ is divisible by 9 if and only if the sum of its digits is divisible by 9. The sum of the digits of $y$ is $250\cdot(2+7+7+2)=250(18)$, so $y$ is divisible by 9.

A number is divisible by 11 if and only if the difference of the sum of the odd numbered digits (the first digit, the third digit, ...) and the sum of its even numbered digits is divisible by 11. The sum of the odd numbered digits of $y$ is $250\cdot(2+7)$ and the sum of the even numbered digits of $y$ $250\cdot(7+2)$. The difference between these two quantities is $0$; so $y$ is divisible by 11.

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This explains it very well! Thank you so much. –  SNS Feb 27 '12 at 23:42

$2772=99 \times 28$

so

$277227722772\ldots277227722772 = 99 \times 28 \times 100010001\ldots000100010001$

and so is divisible by both $9$ and $11$ (and $4$ and $7$ and other numbers).

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It's divisible by $9$, $11$, $4$, $7$, and by every divisor of $100010001 \ldots 000100010001$. Just what the divisors of that last number are is not so easy to tell, as far as I can see. It is not divisible by $4$ or $9$, nor by $2$ or $3$. Whether it's divisible by $7$ or by $11$ might be harder to tell. –  Michael Hardy Feb 28 '12 at 0:14
@Michael: $(10^{1000}-1)/(10^4 -1) = 41 \times 73 \times 137 \times 251 \times 271 \times 401 \times 751 \times 1201 \times 1601 \times 3541 \times 4001 \times 5051 \times 9091 \times 21001 \times 21401 \times 24001 \times 25601 \times 27961 \times 60101 \times 76001 \times 162251 \times 1 378001 \times 1 610501 \times 1 676321 \times 7 019801 \times 1797 655751 \times 5964 848081 \times 10893 295001 \times 182521 213001 \times 14 103673 319201 \times 78 875943 472201 \times 176 144543 406001 \times 1680 588011 350901 \times \cdots$ –  Henry Feb 28 '12 at 0:35

The way you were probably intended to do this problem is to find the sum of the digits (for $9$) and alternating sum and difference (for $11$). And you will undoubtedly need to know these facts about $9$ and $11$ for other problems.

However, the following is true. Suppose that the number formed by a string of digits, like $4718$, is divisible by $m$. For example, $7$ divides $4718$, so let's take $m=7$.

Then $m$ (that is, $7$ here) divides $471847184718$. This is because $$4718471847184718=4718+47180000+471800000000+4718000000000000,$$ and each term on the right-hand side is obviously divisible by $7$, since $4718$ is.

The same argument works for any repetition of the string $4718$, however long it may be, and for any string, and any divisor $m$.

In particular, since $9$ and $11$ each divide $2772$, it follows that each of them divides your thousand digit number. Note that $14$ also divides $2772$, so $14$ divides your thousand digit number.

Your answers were correct, and the procedure that you used turns out to be generally valid. There was somewhat of a lack of explanation, and it is possible that someone grading your work might call it incomplete.

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On the one hand, the divisibility test for $9$ is to add up the digits and see if they're divisible by $9$. The divisibility test for $11$ is a bit more annoying (if combined with $7$ and $14$)- it comes from recognizing that $1001 = 7 \cdot 11 \cdot 13$, so you take the alternating sum of triplets of digits and see if it's divisible by $11$. Or you could just take the regular alternating sum of digits and see if it's divisible by $11$. Both work here.

On the other hand, the number is $\sum_{k = 0}^{249} (1000)^k \cdot 2772$. So any number that divides $2772$ will divide this new number.

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Your reasoning is correct, but you need to find a way to express clearly and justify that last step, from ‘$2772$ is divisible by $9$ and $11$’ to ‘$27722772\dots2772$ is divisible by $9$ and $11$’. I agree that this is intuitively obvious to anyone who has done long division, but ‘intuitively obvious’ isn’t good enough here. Look at a simpler example first: what about $27722772$?

$27722772=27720000+2772=2772\cdot10^4+2772=2772(10^4+1)$, so it’s clearly a multiple of $2772$ and therefore also of $9$ and $11$. Similarly,

\begin{align*} 277227722772&=277227720000+2772\\ &=2772(10^4+1)\cdot 10^4+2772\\ &=2772(10^{2\cdot 4}+10^4)+2772\\ &=2772(10^{2\cdot 4}+10^4+1)\;, \end{align*}

which is again a multiple of $2772$ and hence also of $9$ and $11$.

And if you already know that a string of $n$ copies of $2772$ is equal to

$$2772(10^{(n-1)\cdot 4}+10^{(n-2)\cdot4}+\dots+10^4+1)\;,$$

then a string of $n+1$ copies must be equal to

\begin{align*} \Big(2772(10^{(n-1)\cdot 4}&+10^{(n-2)\cdot 4}+\dots+10^4+1)\Big)\cdot 10^4+2772\\ &=2772(20^{n\cdot 4}+10^{(n-1)\cdot 4}+\dots+10^{2\cdot4}+10^4)+2772\\ &=2772(20^{n\cdot 4}+10^{(n-1)\cdot 4}+\dots+10^{2\cdot4}+10^4+1)\;, \end{align*}

and the pattern continues. Thus, your string of $250$ copies of $2772$ must be equal to

$$2772(10^{249\cdot4}+10^{249\cdot4}+\dots+10^4+1)$$

and is certainly a multiple of $2772$ and therefore of $9$ and of $11$.

Note that this argument actually does more than is necessary, since it also determines the other factor of your number, namely,

$$10^{249\cdot4}+10^{249\cdot4}+\dots+10^4+1=\frac{10^{250\cdot 4}-1}{10000-1}=\frac{10^{1000}-1}{9999}\;.$$

You could use the same kind of reasoning to argue that tacking a copy of $2772$ on the end of some integer $n$ is simply replacing $n$ by $10000n+2772$, so if $n=2772m$, then $10000n+2772=10000(2772m)+2772=2772(10000m+1)$, which is certainly still a multiple of $2772$. Thus, since tacking a copy of $2772$ on the end always preserves divisibility by $2772$ if it was already present, it doesn’t matter how often you do it: the result is still divisible by $2772$ if the original number was.

(Technically both of these arguments are proofs by mathematical induction, albeit stated rather informally.)

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