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?
 A: $2772=99 \times 28$ 
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).   
A: 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. 
A: 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.
A: 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.
