Divisible By 41

It isn't very elegant, but just doing 11/41, 111/41,... you get 11111/41 = 5

I used the method to multiply of elementary school student. I'm new here, and I don't find a how to present this kind of method.

1. You see that for the result to be all one, the first number must be 1. So you have the upper line 1, and the second line 4x1=4. Pic

2. For the result to be 1, the upper line must be 7 Pic Therefore the next number on the second line is 8, because 4x7 = 28 (write 8, keep 2)

3. Remember that your previous calculation, 4+7 =11, not 1, so you must keep 1. Therefore the number on the first line is 2 (2 + 8 = 10, and add your 1 is 11. Write 1, keep 1) Pic

4. 4x2=8, and add 2 from the 4x7=28 calculation, you have 10. Write 0, keep 1. Remember the previous calculation, 2+8+1=11, you kept 1, so you have to make the sums of the row to 10 to make 10 +1 =11. However, you can't do that with a 0 on the bottom line. The upper line therefore will be 0 too, so 0 + 0+1=1 Pic

5. 4x0=0, and add 1 from the 4x2+2 = 10, you have 1 on 2nd line. Aware that you only did 0+0+1 =1 in the previous calculation, and no keeping. Thus, you have to add up to 11 or 1 this time. You can't add to 11, so you add 0 to make 0+1=1 Pic

6. 4x0=0, and no keeping, therefore it's 0 on the bottom line. Hence the upper line is 1. Pic

7. Notice that it has come back to 1. It's easy now. So you find out the number is ...271002710027100...00271. Do whatever you want to find out the answer. Use a calculator, or prove anything. I used a calculator and count the number for my dinner's sake.

I just used long division to see if there is a pattern and found that $41 \times 271 = 11111$ .

Every time 5 ones come together the number is divisible by 41!!!

A JAVA code is all you need

Here is another solution 1111111111/41=27100271 .Therefore answer. also can be 10

11..1(n times) is divisible by 41=>41| 99..9(n times).now 99..9=(10)^n-1. now order of 10 modulo 41 must divide n which is 5

when we take that n 1's are divisible by 41 then by checking we get that after every five number a integer as a result , than n must be factor of five.

The smallest number is 11111 in which n=5 so the largest factor of 5 is 5 itself