### 17

Divisibility by 17

Method 1:

The number is divisible by 17 if the number follows the following method and the remainder is zero.

Let us find the remainder of 42653876 when divided by 17.
Remember the sequence 1, 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12, 1, 10, 15, 14, 4....
Here 1, 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12 repeat in the sequence.
0.Reverse the order of
the digits
6 7 8 3 5 6 2 4
1.Multiply with
the sequence
6×1 7×10 8×15 3×14 5×4 6×6 2×9 4×5
-- 6 70 120 42 20 36 18 20
3.Repeat step 0 with
2 3 3 0 0 0 0 0
4.Multiply with
the sequence
2×1 3×10 3×15 0 0 0 0 0
-- 2 30 45 0 0 0 0 0
2+30+45 =77

7.Divide the result by 17
find the remainder
mod(77/17)=9
If the remainder is zero (0) then the number is divisible by 17 else the result is remainder.

The remainder is 9. Repeat the step 0,1 and 2 till you get a number whose remainder you can find easily.

Method 2:

The number is divisible by 17 if the number follows the following method and the remainder is zero.

Let us find the remainder of 42653876 when divided by 17.
Remember the sequence 1, -7, -2, -3, 4, 6, 9, 5, -1, 7, 2, 3, -4, -6, -9, -5, 1, -7, -2, -3, 4,....
Here 1, -7, -2, -3, 4, 6, 9, 5, -1, 7, 2, 3, -4, -6, -9, -5 repeat in the sequence. Here 1, -7, -2, -3, 4, 6, 9, 5 is followed by negative of them in order.
0.Reverse the order of
digits
6 7 8 3 5 6 2 4
1.Multiply with
the sequence
6×1 7×-7 8×-2 3×-3 5×4 6×6 2×9 4×5
-- 6 -49 -16 -9 20 36 18 20