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by anshumishra » Thu Dec 30, 2010 2:09 pm
Night reader wrote:Which number is divisible by 11?

(I) 1937684514
(II) 2314151667
(III) 1415171802

A. III only
B. I and II
C. II only
D. II and III
E. I, II, III
A

Used this rule : copy/paste from a resource : (By the way, one can do that it using a pen/paper - shouldn't take long).

Alternately add and subtract the digits from left to right. (You can think of the first digit as being 'added' to zero.)
If the result (including 0) is divisible by 11, the number is also.
Example: to see whether 365167484 is divisible by 11, start by subtracting:
[0+]3-6+5-1+6-7+4-8+4 = 0; therefore 365167484 is divisible by 11.


1937684514 -> 0+1-9+3-7+6-8+4-5+1-4 = 15-33 = -18 -> Not divisible by 11
2314151667 -> 0+2-3+1-4+1-5+1-6+6-7 = 11-25 = -14 -> Not divisible by 11
1415171802 -> 0+1-4+1-5+1-7+1-8+0-2 = 4-26 = -22 -> divisible by 11.
Last edited by anshumishra on Thu Dec 30, 2010 2:27 pm, edited 1 time in total.
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by Night reader » Thu Dec 30, 2010 2:14 pm
anshumishra wrote:
Night reader wrote:Which number is divisible by 11?

(I) 1937684514
(II) 2314151667
(III) 1415171802

A. III only
B. I and II
C. II only
D. II and III
E. I, II, III
E

Used this rule : copy/paste from a resource : (By the way, one can do that it using a pen/paper - shouldn't take long).

Alternately add and subtract the digits from left to right. (You can think of the first digit as being 'added' to zero.)
If the result (including 0) is divisible by 11, the number is also.
Example: to see whether 365167484 is divisible by 11, start by subtracting:
[0+]3-6+5-1+6-7+4-8+4 = 0; therefore 365167484 is divisible by 11.
Anshu :( your rule is perfect, now let's get to executing new rule
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by anshumishra » Thu Dec 30, 2010 2:22 pm
Night reader wrote:
anshumishra wrote:
Night reader wrote:Which number is divisible by 11?

(I) 1937684514
(II) 2314151667
(III) 1415171802

A. III only
B. I and II
C. II only
D. II and III
E. I, II, III
E

Used this rule : copy/paste from a resource : (By the way, one can do that it using a pen/paper - shouldn't take long).

Alternately add and subtract the digits from left to right. (You can think of the first digit as being 'added' to zero.)
If the result (including 0) is divisible by 11, the number is also.
Example: to see whether 365167484 is divisible by 11, start by subtracting:
[0+]3-6+5-1+6-7+4-8+4 = 0; therefore 365167484 is divisible by 11.
Anshu :( your rule is perfect, now let's get to executing new rule
Oops !
It is A.

1937684514 -> 0+1-9+3-7+6-8+4-5+1-4 = 15-33 = -18 -> Not divisible by 11
2314151667 -> 0+2-3+1-4+1-5+1-6+6-7 = 11-25 = -14 -> Not divisible by 11
1415171802 -> 0+1-4+1-5+1-7+1-8+0-2 = 4-26 = -22 -> divisible by 11.
Thanks
Anshu

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by gmat7202011 » Thu Dec 30, 2010 7:02 pm
Anshu,

Can you please shed some more light on this rule please. Is this applicable for all prime numbers ? What scenarios can i apply this rule.

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by anshumishra » Thu Dec 30, 2010 7:20 pm
gmat7202011 wrote:Anshu,

Can you please shed some more light on this rule please. Is this applicable for all prime numbers ? What scenarios can i apply this rule.

Thank You
gmat7202011,

This rule is just for testing divisibility by 11. I don't think you necessarily need to know this rule though.
Here it is mentioned : https://mathforum.org/dr.math/faq/faq.divisibility.html

I would worry about knowing it for 2,3,4,5,6,8,9.

This has a generic rule : https://mathforum.org/dr.math/faq/faq.divisibleto50.html
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by jwortma2 » Thu Dec 30, 2010 8:23 pm
anshumishra wrote:
Night reader wrote:Which number is divisible by 11?

(I) 1937684514
(II) 2314151667
(III) 1415171802

A. III only
B. I and II
C. II only
D. II and III
E. I, II, III
A

Used this rule : copy/paste from a resource : (By the way, one can do that it using a pen/paper - shouldn't take long).

Alternately add and subtract the digits from left to right. (You can think of the first digit as being 'added' to zero.)
If the result (including 0) is divisible by 11, the number is also.
Example: to see whether 365167484 is divisible by 11, start by subtracting:
[0+]3-6+5-1+6-7+4-8+4 = 0; therefore 365167484 is divisible by 11.


1937684514 -> 0+1-9+3-7+6-8+4-5+1-4 = 15-33 = -18 -> Not divisible by 11
2314151667 -> 0+2-3+1-4+1-5+1-6+6-7 = 11-25 = -14 -> Not divisible by 11
1415171802 -> 0+1-4+1-5+1-7+1-8+0-2 = 4-26 = -22 -> divisible by 11.
Can you please explain where you got the numbers which are in bold above?
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by anshumishra » Thu Dec 30, 2010 8:56 pm
jwortma2 wrote:
anshumishra wrote:
Night reader wrote:Which number is divisible by 11?

(I) 1937684514
(II) 2314151667
(III) 1415171802

A. III only
B. I and II
C. II only
D. II and III
E. I, II, III
A

Used this rule : copy/paste from a resource : (By the way, one can do that it using a pen/paper - shouldn't take long).

Alternately add and subtract the digits from left to right. (You can think of the first digit as being 'added' to zero.)
If the result (including 0) is divisible by 11, the number is also.
Example: to see whether 365167484 is divisible by 11, start by subtracting:
[0+]3-6+5-1+6-7+4-8+4 = 0; therefore 365167484 is divisible by 11.


1937684514 -> 0+1-9+3-7+6-8+4-5+1-4 = 15-33 = -18 -> Not divisible by 11
2314151667 -> 0+2-3+1-4+1-5+1-6+6-7 = 11-25 = -14 -> Not divisible by 11
1415171802 -> 0+1-4+1-5+1-7+1-8+0-2 = 4-26 = -22 -> divisible by 11.
Can you please explain where you got the numbers which are in bold above?
0+1-9+3-7+6-8+4-5+1-4 = (1+3+6+4+1)-(9+7+8+5+4) = 15-33 = -18.
0+2-3+1-4+1-5+1-6+6-7 = (2+1+1+1+6)-(3+4+5+6+7) = 11-25 = -14
0+1-4+1-5+1-7+1-8+0-2 = (1+1+1+1+0)-(4+5+7+8+2) = 4-26 = -22
Thanks
Anshu

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by Anurag@Gurome » Thu Dec 30, 2010 10:32 pm
To put Anshu's method in simple words: If the difference of the sum of even numbered digits and the sum of odd number digits is divisible by 11, then the number is divisible by 11.

For example take option II and option III,
  • I. 2314151667
    • Sum of even numbered digits = (3 + 4+ 5 + 6 + 7) = 25
      Sum of odd numbered digits = (2 + 1 + 1 + 1 + 6) = 11
      Their difference = (25 - 11) = 14 => Not divisible by 11
    II. 1415171802
    • Sum of even numbered digits = (4 + 5 + 7 + 8 + 2) = 26
      Sum of odd numbered digits = (1 + 1 + 1 + 1 + 0) = 4
      Their difference = (26 - 4) = 22 => Divisible by 11
Note: Even (or odd) numbered digit means even (or odd) positioned digits. And there should not be any confusion whether to start numbering from left or right as we are taking the difference of the sums.
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