A number is divisible by ……..By 7, 11, and 13 if it consists of but four places, the first and fourth being occupied by the same
significant figures, and the second and third by ciphers.”
(A cipher is a 0.) It continues:
“Thus, 2002, 3003, and 5005 are divisible by 7, 11, and 13.”

Proof:  7×11x13 = 1001Similar Posts:

• Divisibility
• DIVISIBILITY RULE FOR 7
• Use Negative Remainders

(1) Subtract the ones digit from N: Subtract 3 from 273: 273
– 3
270
(2) Dividing the result by 10: Divide 270 by 10: 270 ÷ 10 = 27
(3) Subtract—from that result—twice
the original ones digit: Subtract, from 27, twice 3:

27

– 6

21:  a multiple of 7
Since this procedure leads to a multiple of 7, it must be that the original number, 273, is a
multiple of 7, as shown in this long division:
A much quicker way to use the procedure is shown here
2 7 3
– 6
2 1
x 2
Since 21 is a multiple of 7, it must
be that 273 is a multiple of 7.

Let’s apply this more direct procedure to a larger multiple of 7: 7 x 568 = 3,976:
3 9 7 6
– 1 2
3 8 5
x 2
We don’t know if 385 is a multiple of 7
or not, so we continue the procedure.
3 8 5
– 1 0
2 8

Since 28 is a multiple of 7, it must be that
both 385 and 3,976 are multiples of 7.
What is the proof behind this procedure?
Consider, without loss of generality, a three digit number, N = 100a + 10b + c, where a, b, and c are
single digits, a ≠ 0.
Applying the Divisibility Rule for 7 to 100a + 10b + c, we get:
(1) Subtract c: 100a + 10b
(2) Divide by 10: 10a + b
(3) Subtract twice c: 10a + b – 2c
Claim: 100a + 10b + c is a multiple of 7 if and only if 10a + b – 2c is a multiple of 7
PSimilar Posts:

• Divisibility
• Use Negative Remainders
• Divisible by 7, 11 and 13

First of all, What is Negative Remainder:

eg. When 14 is divided by 8

Then, 14= 8×1+6

or

14=8×2-2

Therefore, We have two remainder and they are +6 and -2 where (-2) is the Negative Remainder. I hope I am able to make it clear.

Now, I make a small assumption that you know the remainder theorum and you make use of it while calculating the remainder. Ok, Now lets take an example

Q. Find the remainder when 14×15 is divided by 8

Ans. Traditional method: Find remainder after dividing 14 and 15 by 8, multiply them and find the final answer i.e

a)14/8= 8×1+6

b)15/8=8×1+7

c)find remainder after dividing 6×7 by 8 = 42/8=8×5+2

Therefore, Remainder is 2

New method,

a)14=8×2-2 (minus 2 is the negative remainder)

b)15=8×2-1 ( minus 1 is the negative remainder)

Final Remainder is = (-2)x(-1) divided by 8 =2

Therefore Remainder is 2.

Though both the method looks same in calculation level but when you some difficult problem then this negative remainder method will help a lot and will save some time.s

Remember: If at the end remainder is negative then substract that number from the divisor for eg. If the answer was (-3) then remainder would be = 8-3=5.Similar Posts:

• CALENDER
• Square of a two digit number
• DIVISIBILITY RULE FOR 7
• Easy Way to Multiply two numbers
• Divisibility

Any six-digit, or twelve-digit, or eighteen-digit, or any such number with number of digits equal to multiple of 6, is divisible by EACH of 7, 11 and 13 if all of its digits are same .

For example 666666, 888888888888 etc. are all divisible by 7, 11, and 13. Similar Posts:

• DIVISIBILITY RULE FOR 7
• Divisible by 7, 11 and 13
• Use Negative Remainders