Without calculator and excel

[Night reader]

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

[anshumishra]

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.

[Night reader]

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

[anshumishra]

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.

[gmat7202011]

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

[anshumishra]

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

[jwortma2]

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?

[anshumishra]

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

[Anurag@Gurome]

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.