Rule of divisibility by 7.?

Question

I want the rule for divisibilty by 7 as well as the proof? Apply that rule to check if for example 398258 is divisible by 7 or not.

there is a rule for this..

there is a rule for every division till 13 ..or so..

blahb31..you are right..can u give me the proof..also nimmi..ur rule is not right..just check on 63..e.g..

rule is fine..but the proof for this..i want that..

Answer 1

There are actually a lot of divisibility tests for the number 7. I'll prove the one that's been mentioned the most so far in the answers above, which is: "take off the last digit, double it, and subtract that result from the rest of the number."

For instance: is 1064 divisible by 7? Well:

106 - (2*4) = 106-8 = 98. But is 98 divisible by 7?

9 - (2*8) = 9-16 = -7, which is divisible by 7.

But why does it work?

Well, consider that any number of more than two digits can be written as 10a + b, where b is the last digit and a is the number formed from the rest of the digits. (For example, 1064 = 10*106 + 4. a=106, b=4).

The "re-formed" number, then, when we subtract double the last digit, is a - 2b. With me so far?

Now, if a-2b is a multiple of 7, then we can write it as 7k, where k is some integer:

a - 2b = 7k

Now add 49a+7b to both sides:

a - 2b + 49a + 7b = 7k + 49a + 7b

Simplifying the left side:

50a + 5b = 7k + 49a + 7b

Factoring a 5 from the left side, and a 7 from the right:

5(10a + b) = 7(k + 7a + b)

Now, a, b, & k are all integers, so k+7a+b is too. That means the right side is 7 times some integer -- in other words, a multiple of 7.

But the left side is 5 times our original number (10a+b), and since 5 isn't a multiple of 7, 10a+b must be.

That proves, then, that as long as a-2b is a multiple of 7, so is 10a+b.

Hope that helps!

Answer 2

. To find out if a number is divisible by 7, double the last digit, then subtract it from the remaining digits of the number. If you get an answer divisible by 7, then the original number is divisible by 7. If you don't know whether the new number is divisible by 7, you apply the rule again. For example, to check whether 616 is divisible by 7, double the last digit (6 x 2 = 12), then subtract the answer from the remaining digits (61 – 12 = 49). Because 49 is divisible by 7, so is 616

Answer 3

Rule for Divisibility by 7:

Method A:

Use short division left to right,
check remainder for zero.
This means subtracting multiples of 7
from the highest two digits
and repeating this until result is less than 7.
If result is zero, number is divisible by 7.

Method B:

Keep subtracting multiples of 1001
until you get a 2 or 3 digit result.
Check this for divisibility by 7.

Proof:

A is obvious.

B works because 1001 = 7 * 11 * 13.

Example:

1234567-->534567-->44567-->2567-->467->47->5

or

1234567->233567->33367->3337->334->54->5

So 1234567 is NOT divisible by 7.

-- davar55

Answer 4

blahb31 is correct the method is to cut of the last digit, double it subtract it from the remaining number, repeat until simple:
so in your case
398258 --> 39825-16[8*2]=39809
39809 --> 3980 - 18 = 3962
3962-->396-4=392
392 -> 39 -4 = 35
3 - 10 =-7
So 398258 is divisible by 7.

Proof Sketch.
Say your number is x, we can represent it as YZ.. where Y is all decimal digits of a number except for the last one, and Z is the last one.
So if x= 235 then Y= and Z=5.
if x=398258 then Y=39825 and Z=8.

Now, we know that YZ=Y*10+Z, now we subtract 2*Z from Y we get Y-2*Z we need to prove that if YZ is divisible by 7 then so is Y-2*Z.

Let's assume that YZ is divisible by 7. Then so is Y*10+Z.
Now note that Y*3+Z is also divisible by 7, just subtract 7Y. -4*Y+Z is is also divisible, subtract 7Y.

Multiply by -1. get that 4*Y-Z is also divisible by 7.
Multiply by 2, 8*Y-2*Z is also divisible,
subtract 7*Y, we get that Y-2*Z is ALSO divisible by 7 if YZ was.

QED

Answer 5

I have one of them. Add all the digits and if the total is exactly divisible by 7, then the no. is also divisible by 7. I read this somewhere but I'm not sure.
3+9+8+2+5+8= 35/7=5. Now the total of all the digits are exactly divisible by 7 and so this is divisible by 7. Check it yourself.

Answer 6

Rule of divisibility by 7. Multiples of 7 are divisible by 7. 398258 divided by 7 is: 56894

Answer 7

Rule of divisibility by 7:

Double the digit in the ones column, and subtract that answer from the rest of the number without the ones column. If that can be divided by 7, so can the whole number. Repeat as necesary.

Check 398258:

39825 - (2x8) = 39809
3980 - (2x9) = 3962
396 - (2x2) = 392
39 - (2x2) = 35

7 x 5 is 35, so 398258 is divisible by 7.

Good luck!

Answer 8

"To determine if a number is divisible by 7, take the last digit off the number, double it, and subtract the doubled number from the remaining number. If the result is evenly divisible by 7 (e.g. 14, 7), then the number is divisible by seven. This may need to be repeated several times."

Proof is also below.

Answer 9

There is no rule of divisibility for 7. There are multiples of 7, but 398258 does not divide evenly by 7.

Answer 10

Here's an easy proof......notice that any number with 2 digits (or more) can always be written as 10a + b.....for instance 506 = 10(50) + 6 where a = 50 and b = 6

Then let our number be 10a + b ........now, subtract 2b from a and assume that this is divisible by 7.....that is.....a - 2b = 7k where k is an integer.......rearranging, we have......a = 7k + 2b......and substituting this back into 10a + b, we have ..... 10(7k + 2b) + b = 70k + 20b + b = 70K + 21b = 7(10k + 3b)......and this is clearly divisible by 7 !!!!