Div divis divisible

Start with an integer $N$ with a thousand digits, with last digit be $b_1$ , and have the rest of the digits form the number $a_1$ , and so $a_1>b_1$ . We know that, where $k$ is an integer, $\dfrac{10a_1+b_1}{a_1}=k$ $10a_1+b_1=a_1k$ $a_1(k-10)=b_1$ However, we know $a_1>b_1$ , so this only has a solution where $k=10$ and $b_1=0$ . Now, let $a_1=10a_2+b_2$ . Deleting the digit $b_2$ gives, with another $k$ , $\dfrac{100a_2+10b_2}{10a_2}=k$ $\dfrac{10a_2+b_2}{a_2}=k$ $a_2(k-10)=b_2$ We have an identical equation from before, and so $b_2=0$ . This logic will continue to hold until there are two nonzero digits left, and so it is not necessarily true that $a>b$ . Now, similarly, $\dfrac{10a_{1000}+b_{1000}}{a_{1000}}=k$ $a_{1000}(k-10)=b_{1000}$ Finally, we can have $b>a$ , and each pair of digits such that the second is a multiple of the first gives a solution. Each of these pairs can be made into a desired 1000-digit number by appending 998 0's to the end.

The answer is $9+9+4+3+2+1+1+1+1+1=\boxed{32}$ .

First let us note that the 9 choices of $A$ with one non-zero digit followed by 999 zeroes all satisfy the condition.

Now let us assme $A$ consists of more than one non-zero digit. In this case we may write $A=N*10^{k+1}+b*10^{k}$ , where $k$ is the number of terminal zeroes, $N$ is some integer, and $b\leq1\leq9$ is the last non-zero digit. For example if $A=12345000\dots0$ then $N=1234$ and $b=5$ . (By assumption $N$ has at least one digit.)

Now letting $B=N*10^{k}$ be the integer formed by removing $b$ , then since $B$ divides $A$ we can deduce that $N$ divides $b$ . Since $b\leq1\leq9$ and $N$ is a factor of $b$ then $N$ is necessarily a one-digit number.

We hence must have that $A$ has only two non-zero digits with the second digit a multiple of the first. We can easily check that all numbers of this form work.

Combining both type of solution (one and two non-zero digits) we have 10 possibilities starting with 1 (10,11,...19 then all zeroes), 5 starting with 2 (20,22,24,26,28 then all zeroes), 4 starting with 3 (30,33,36,39 then all zeroes), 3 starting with 4 (40,44,48 then zeroes) and 2 starting with 5,6,7,8 and 9 (e.g. 90 and 99 then zeroes). This gives 32 solutions in total.

Note that the answer is the same for any length of A greater than or equal to 2 (since all but the first two digits are zero). Tthe 1000 digits is a red herring.

Let the $1000$ digit number be $A = \overline{a_1 a_2 a_3 \ldots}$ . Suppose we delete $a_2$ . Then, we have $B = \overline{a_1 a_3 a_4 \ldots}$ . Writing these as powers, we have $\frac{A}{B} = \frac{a_1 \times 10^{1000} + a_2 \times 10^{999} + a_3 \times 10^{998} + \ldots}{a_1 \times 10^{999} + a_3 \times 10^{998} + a_4 \times 10^{997} + \ldots}$ We can take out 1 from the fraction: $\frac{A}{B} = 1 + \frac{9a_1 \times 10^{999} + a_2 \times 10^{999}}{a_1 \times 10^{999} + a_3 \times 10^{998} + a_4 \times 10^{997} + \ldots}$ For this fraction to be a whole number (thereby showing that $A$ is divisible by $B$ ), it is apparent that we need $a_3 = a_4 = a_5 = \ldots = a_{1000} = 0$ . Suppose otherwise. The numerator then will have $10^{999}$ as a factor while the denominator will not - thus the fraction cannot be a whole number.

With this in mind, we can rewrite our fraction: $\frac{A}{B} = 1 + \frac{9a_1 \times 10^{999} + a_2 \times 10^{999}}{a_1 \times 10^{999}}$ It then follows $\frac{A}{B} = 1 + \frac{9a_1 + a_2}{a_1}$

The number of solutions to this is simply, as $a_1$ varies from $1$ through $9$ , the number of possible values of $a_2$ there are such that $a_2$ is divisible by $a_1$ . Note that $a_2$ ranges from $0$ through $9$ . Simple casework shows this to be $10 + 5 + 4 + 3 + 2 + 2 + 2 + 2 + 2 = \boxed{32}$ .

Let $F[n]$ be the total number of N-digit numbers A satisfying that when any digit of A, aside from the first, is deleted to form a N-1 digit number B, then B divides A.

At first, we pay attention to all 2-digit numbers satisfying this condition. There are 32 numbers: $10, 11, 12, .... ,90, 99.$ With a number A satisfying the condition, there is only one way to create a 3-digit number X has this property by adding digit 0 after A. It implies that the number of 2-digit numbers and the number of 3-digit ones have this property are equal.

Proof by induction, we have the result that $F[2] = F[3] = ... = F[n] = ...$ for all $n \ge 2.$

Therefore, the answer is 32.

For any $N$ of $n>2$ digits, the only number divisible by $N$ that has the same first digit is $\frac{N}{10}$ . Thus, easily proved by induction, we know that any $n$ -digit number must be divisible by $10^{n-2}$ , i.e., two digits followed by $n-2$ zeros.

This means that this combinatorial value holds for any amount of digits, being equal to the amount for just two digits. The two-digit numbers with this property are 10-19, 20,22,24,26,28, 30,33,36,39,40,44,48,50,55,60,66,70,77,80,88,90,99, for a total of $\boxed{32}$ possibilities.

Let us find all k+1-digit numbers of this sort, for all k\geq 1.

If k=1, then we must delete the last digit. This leaves a 1-digit number. The whole number must be divisible by this digit; that means that the deleted digit must also be divisible by this digit. Thus, the possibilities are any number starting with 1, any even number starting with 2, any number divisible by 3 starting with 3, and so on. So the choices are

10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 24, 26, 28, 30, 33, 36, 39, 40, 44, 48, 50, 55, 60, 66, 70, 77, 80, 88, 90, 9...

giving 32 numbers.

Now assume k\geq 2, and let us suppose we delete the last digit to get a number c. Then c|10c+d where d is a digit; we get from this that c|d, but c>d so d=0. Thus the number ends in 0. If we delete any other digit, we get the statement that some number ending in 0 divides some other number ending in 0, each of which we can divide by 10. This is equivalent to the k-digit number c having the same property.

By induction we can claim that the only such numbers are the 32 listed above times a correct power of 10; the base case is k=1 and the inductive step we have proven above. So there are only 32 such 1000-digit numbers.

The numbers are: 10 000 ..., 11 000 ..., ... , 19 000 ... ; 20 000 ..., 22 000 ..., 24 000 ..., 26 000 ..., 28 000 ...; 30 000 ..., 33 000 ..., 36 000 ..., 39 000 ...; 40 000 ..., 44 000 ... 48 000 ...; 50 000 ..., 55 000 ...; 60 000 ...; 66 000 ...; 70 000 ..., 77 000 ...; 80 000 ...; 88 000 ...; 90 000 ..., 99 000 ... . "1000 digits" no matter here, only first two digits (number of final zero-es could be as much as you want) !!!

Let us first find out the number of two digit numbers satisfying the above property. Let the two digit number be $ab$ . Now we know that $ab=10\times a + b$ . This number must be divisible by $a$ (because it satisfies the above property). Therefore,

$10\times a + b = n\times a \implies b=a\times (n-10)$ . Since $b$ is a non zero integer, $n \ge 10$

Consider cases now.

Case 1 : $n=10$ . This implies that $b=0$ and $a$ can take all values from $1$ to $9$ . Hence 9 such numbers are possible.

Case 2 : $n=11$ . This implies that $b=a$ and $a$ can take all values from $1$ to $9$ . Hence 9 such numbers are possible.

Case 3 : $n=12$ . This implies that $b=2 \times a$ and $a$ can take all values from $1$ to $4$ . Hence 4 such numbers are possible.

Performing similar operations till n=19 gives us all possibile two digit numbers satisfying above property. Hence total is $9+9+4+3+2+1+1+1+1+1=32$ .

Now consider a three digit number $abc$ which satisfies above property. Now we can write:

$abc = ab\times n \implies c = (10\times a + b)\times (n-10)$ ...eq $1$ and

$abc = ac\times m$ ...eq $2$ .

Since $0 \le c \le 9$ integer and $(10\times a+b)$ is a two digit number, the only value that n can take is 10 and hence $c=0$ . Substituting $c=0$ in eq $2$ gives:

$b=a\times (m-10)$ which is nothing but the same as the two digit analysis we did above. Hence number of three digit numbers that satisfy the above condition are $32$ . By the same logic, the number of 1000 digit numbers $A$ satisfying the above property $=32$ . .

Let $A_k$ be the number obtained by deleting the $k$ th digit of $A$ . Let $X_k$ be the $k$ -digit number obtained by taking the first $k$ digits of $A$ . Let $Y_k$ be the $k$ -digit number obtained by taking the last $k$ digits of $A$ . Then (defining $Y_0=0$ ) $A_j = 10^{1000-j}X_{j-1} + Y_{1000-j} \qquad A = 10^{1001-j}X_{j-1} + Y_{1001-j}$ for $2 \le j\le 1000$ . Hence $A - 10A_j = Y_{1001-j} - 10Y_{1000-j} \qquad 2 \le j \le 1000$ Since $A_j$ divides $A$ for all $2 \le j \le 1000$ , we deduce that $A_{1000-k}$ divides $Y_{k+1} - 10Y_k$ for all $0 \le k \le 998$ . Since $Y_k$ is a $k$ -digit number, we see that $|Y_{k+1} - 10Y_k| < 10^{k+1}$ for all $0 \le k \le 998$ . Since $A_{1000-k}$ is a $999$ -digit number, and hence at least $10^{998}$ , we deduce that $Y_{k+1}-10Y_k = 0$ for all $0 \le k \le 997$ . Thus $Y_k=0$ for all $0 \le k \le 998$ , and hence the last $998$ digits of $A$ must be zero.

Thus $A$ must be of the form $ab0000\cdots 000$ . We have not yet used the condition that $A_2$ divides $A$ . This tells us that $a$ divides $ab$ . The $32$ choices for $ab$ are:

10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 24, 26, 28, 30, 33, 36, 39, 40, 44, 48, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95.

Consider N. Let b be the last digit (lowest value). N can be rewritten as 10xA+b. After deletion, we have number A that needs to divide 10xA+b. The remainder of this division is b. So A needs to divide b. There are two cases: either b is zero, or b is a multiple of A and therefore, A must consist of 1 digit at most.

Now, as long as the 'preamble' A consists of 2 or more digits, the same reasoning holds: N can be rewritten as (10xA+b) x 10^n where n equals 1000-length(A)-1. After deletion of b, we have number Ax10^n that needs to divide (10xA+b) x 10^n. The remainder of this division is b x 10^n. So A needs to divide b. Again, there are two cases: either b is zero, or b is a multiple of A.

This reasoning can be repeated (case A > 9, so b must be 0) until we have zeroed out 998 digits. What's left are the cases of A consisting of 1 digit dividing the single digit b (and the cases where b is still zero here): 10,11,12,13,14,15,16,17,18,19,20,22,24,26,28,30,33,36,39,40,44,48,50,55,60,66,70,77,80,88,90,99 where all these 32 'preambles' are followed by the 998 zeroes.