In this post we will discuss about the methods by which we can find the remainder of a number when divided by any other number easily. We know methods for most of the numbers. But how they are found are prime importance to us, as they can be useful to us. The methods described here can be used to find remainders of any number when they are divided by any other number. We will first give some rules to find the remainders for some divisors.
We can follow the similar methods for all the numbers. When the remainders for 10x where x is an integer gives a pattern then the pattern can be used for making the methods simple.
To have a detailed description visit https://sites.google.com/site/rahulsmaths/idiscover/2013/remainder.
- 2
- If the digit at the unit place is zero or multiple of two which is less than 10, i.e. either they are 0,2,4,6 or 8. The remainder of the number is either 0 or 1.
- 3
- If the sum of the digits is divisible by 3 then the number is divisible by 3. The remainder is the number which is left when the process is repeated again with the obtained sum.
- 4
- The number is divisible by 4 if the last two digits is divisible by 4. The remainder is the remainder obtained by the division of the last two digits.
- 5
- If the digit at the unit place is either 0 or 5 then the number is divisible by 5. The remainder is the remainder when the last digit is divided by 5.
- 6
- If the number is divisible by 2 and 3 both then the number is divisible by 6. It can also be checked by multiplying every digit by 4 and adding and repeat the process till you get a number which is easily divisible by 6. If the remainder left is zero then the number is divisible by 6 else the remainder is the remainder.
- 7
- If the digits of the number starting from the unit place is multiplied by the digits 1,3,2,6,4,5 which is repeated for the left digits and all the results are added and the process is repeated till you get a number easily divisible by 7. If the remainder left at the last is zero then the number is divisible by 7 else not. The remainder is the remainder found at last.
- 8
- If the last three digits is divisible by 8 then the number is divisible by 8 else not. The remainder is the remainder found from the last three digits.
- 9
- If the sum of the digits is divisible by 9 then the number is divisible by 9. If the sum is large then the method can be applied to this number also. The remainder is the remainder of the sum.
We can follow the similar methods for all the numbers. When the remainders for 10x where x is an integer gives a pattern then the pattern can be used for making the methods simple.
To have a detailed description visit https://sites.google.com/site/rahulsmaths/idiscover/2013/remainder.
Let us have an example of divisibility by 7.
Divisibility by 7
Remainder
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.
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 | |||||||||
| 6.Divide the result by 7 find the remainder | |||||||||
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.
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.
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 | |||||||||
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.
Repeat the step 0, 1 and 2 till you get a number whose remainder you can find easily.
Remainder
- In method 1 we can surely say that the result is the remainder of the original number.
- 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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