### 21

Divisibility by 21

Method 1:

The number is divisible by 21 if the number follows the following method and the remainder is zero.

Let us find the remainder of 42653876 when divided by 21.
Remember the sequence 1, 10, 16, 13, 4, 19, 1, 10, 16, 13, 4, 19, 1, 10, 16, 13, 4, 19, 1, 10, 16,....
Here 1, 10, 16, 13, 4, 19 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 8×16 3×13 5×4 6×19 2×1 4×10
-- 6 70 128 39 20 114 2 40
3.Repeat step 0 with
9 1 4 0 0 0 0 0
4.Multiply with
the sequence
9×1 1×10 4×16 0 0 0 0 0
-- 9 10 64 0 0 0 0 0
9+10+64 = 83
6.Divide the result by21
find the remainder
mod(83/21)=20
If the remainder is zero (0) then the number is divisible by 21 else the result is remainder.

The remainder is 20.  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 21 if the number follows the following method and the remainder is zero.

Let us find the remainder of 42653876 when divided by 21.
Remember the sequence 1, 10, -5, -8, 4, -2, 1, 10, -5, -8, 4, -2, 1, 10, -5, -8, 4, -2, 1, 10, -5,...
Here 1, 10, -5, -8, 4, -2 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×-5 3×-8 5×4 6×-2 2×1 4×10
-- 6 70 -40 -24 20 -12 2 40