7

Divisibility by 7

Method 1:

The number is divisible by 7 if the number follows the following method and the remainder is zero.

Let us find the remainder of 42653876 when divided by 7.
Remember the sequence 1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, 1,....
Here 1, 3, 2, 6, 4, 5 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 7×3 8×2 3×6 5×4 6×5 2×1 4×3
-- 6 21 16 18 20 30 2 12
2.Add the numbers 6+21+16+18+20+30+2+12=125
3.Repeat step 0 with
the added number
5 2 1 0 0 0 0 0
4.Multiply with
the sequence
5×1 2×3 1×2 0 0 0 0 0
-- 5 6 2 0 0 0 0 0
5.Add all
5+6+2 = 13
6.Divide the result by 7
find the remainder
mod(13/7)=6
If the remainder is zero (0) then the number is divisible by 7 else the result is remainder.

The remainder is 6.  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 7 if the number follows the following method and the remainder is zero.

Let us find the remainder of 42653876 when divided by 7.
Remember the sequence 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1,....
Here 1, 3, 2, -1, -3, -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 7×3 8×2 3×-1 5×-3 6×-2 2×1 4×3
-- 6 21 16 -3 -15 -12 2 12
2.Add the numbers 6+21+16-3-15-12+2+12=27
3.Repeat step 0,1 and 2 with
the result, if number is very large and positive
6.Divide the result by 7
find the remainder
mod(27/7)=6
If the remainder is zero (0) then the number is divisible by 7.

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

Remainder

  1. In method 1 we can surely say that the result is the remainder of the original number.
  2. In method 2 we are not sure for the remainder. If the result is positive then its remainder is the remainder if the result is negative then 7+negative value is the remainder.


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