Without calculator and excel

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.

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

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.

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 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

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?

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