Divided by 11s.... United by remainders

We note that $407 \equiv 0 \mod{11}$

Therefore, $417 = 407+10 \equiv 10 \mod{11}$

Now, $4170 = 4070 +100 = 4070 + 99 + 1 \equiv 1 \mod{11}$

Note that $1$ and $10$ are the smallest and largest remainderx possible and their difference is $10-1=\boxed{9}$ .

The number that will form the largest remainder is 417 (which gives the remainder as 10)and the number that gives the smallest remainder is 4170(a remainder of 1).

Therefore the difference between the largest and the smallest remainders is 9.

10407 when divided by 11 gives 10 as reminder and when 1004070 is divided gives 1. hence 9.

Interestingly, I noted that the divisibility of a number by 11 ties fairly accurately to its remainder when divided by 11. By divisibility rule, for a number to be divisible by 11, the sum of odd digits - sum of even digits = 11. However, it would seem that if the rule number (i.e., the sum of all odd digits -sum of all even digits) were divided by 11, the remainder would be the same. Look at the following cases:

Unique Case: All digits are odd (e.g. 10407)

$11|10407 = 946,$ rem $1$ ( $1+4+7-(0)=12$ (or rem 1))

Now, for any other number, the digits given will have two of them being summed, and the third subtracted from it (e.g. $1+7-4=4$ ). From this we can develop the table below:

Interestingly, you can see that there is a direct link between the last two columns, and you can easily use this principle to determine the largest and smallest remainders of the given numbers. (Of course, my theory needs some tweaking once 0's come in, but you get the picture.)

So from here, we realize that the largest and smallest remainders will be 10 and 1, the differences of which will be 9.

Here we have to prove that no such number is divisible by 11.

With the given conditions we have infinitely many numbers.

1004070, 407000001 etc.

FOR TEST of DIVISIBILITY by 11:-

Let P and Q be the sum of digits at odd places and even places respectively.

Let D be the difference of P and Q.

The possible values of P and Q are:-

If P = 1+4+7=12 then Q = 0 and D = 12.

If P = 1+4 = 5 then Q = 7 and D = 2.

If P = 1+7 = 8 then. Q =4 , D = 4.

If P = 7+4 = 11 then Q = 1, D = 10.

If P = 1 then Q = 7+4 = 11 , D = 10

If P = 4 then Q = 1+7 = 8.

If P = 7 then Q = 1+4 = 5.

(Note: digits 1, 4 & 7 occur only once in a number).

In each case Since D is not divisible by 11 therefore the number is not divisible by 11.

[Consider 1004070. Here digits at first place, third place, fifth place and seventh place are 0, 0, 0 and 1 resp. Digits at 2nd, 4th and 6th places are 7, 4 and 0 res.]

Thus remainder can not be 0.

7041 gives 1 as remainder and 714 gives 10 as remainder.

Hence Ans = 10-1 = 9.

as 407 gives 0 remainder

hence 4071 will give 1 as a remainder

where as 417 gives 10 as a remainder

hence 10-1=9