Latest Accenture Aptitude Question SOLUTION: Which of the following numbers are completely divisible by 11? A. 3245682 B. 283712 C. 438416 D. 36894 1) Only A 2) Only B 3) Only C 4) Only D 5) All are divisible

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(#M40037619) ACCENTURE Numerical Ability question Keep an EYE

Which of the following numbers are completely divisible by 11?
A. 3245682
B. 283712
C. 438416
D. 36894
1) Only A
2) Only B
3) Only C
4) Only D
5) All are divisible

Asked In Accenture Sundar Kumar (9 years ago)

Unsolved Read Solution (43)

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Read Solution (43) :

sudhir  (9 years ago)

all are divisible.....check as if (sum of odd place digit-sum of even place digit)you will get either 0 or 11.hence all number is divisible by 11

Is this solution Helpfull? Yes (31) | No (3)  

Saraswathy  (9 years ago)

For a no. to be divisible by 11, diff. of sum of digits at odd places nd sum of digits at even places should be divisible by 11...
In the given options , Only B satisfies the condition...
283712 => (2+7+8) - (1+3+2) => 17-6 = 11

Ans : 2) Only B

Is this solution Helpfull? Yes (9) | No (23)  

jyoti survase  (9 years ago)

all are divisible by 11,because diff of sum of digits at odd place nd sum of digits at even digit is 0 or divisible by 11

Is this solution Helpfull? Yes (5) | No (2)  

vijeta kumari  (9 years ago)

option 5
because when the difference of sum of odd position number and even position number is divisible by 11 then the number is completely divisible by 11.

Is this solution Helpfull? Yes (4) | No (2)  

YUVA  (9 years ago)

all are divisible

Is this solution Helpfull? Yes (2) | No (1)  

Sundar Kumar  (9 years ago)

Divisibility By 11 : A number is divisible by 11, if the difference of the sum of its
digits at odd places and the sum of its digits at even places, is either 0 or a number
divisible by 11.
Ex. The number 283712 is divisible by 11, since :
(sum of digits at odd places) - (sum of digits at even places)
= (8+ 7 + 2)-(2+ 3 +1) = 11, which is divisible by 11.
Answer: B)283712

Is this solution Helpfull? Yes (2) | No (3)  

nagulmeera  (9 years ago)

3245682/11=29562
283712/11=2579
438416/11=3985
36894/11=3354
so ans is 5

Is this solution Helpfull? Yes (2) | No (1)  

Abhishek Agarwal  (9 years ago)

diff. of sum of digits at odd places nd sum of digits at even places are either 0 or multiple of 11 so in this case we have 0 & 11 option 5 is true

Is this solution Helpfull? Yes (1) | No (2)  

Dhananjay Rajmane  (9 years ago)

3
Sum of odd placed digit-sum of even placed digit should be divisible by 11.
4+8+1=13
3+4+6=13
13-13=0
hence it i divisible by11

Is this solution Helpfull? Yes (1) | No   

Rohini kaushik  (9 years ago)

a no. is divisible by 11 if the difference of sum of digit at odd places and sum of digits at even places is either 0 or 11.
ans- 5

Is this solution Helpfull? Yes (1) | No   

banupriya  (9 years ago)

only B
because add odddigit- addevendigit that number is divisible by 11 means that's the answer

Is this solution Helpfull? Yes (1) | No   

pamesh  (9 years ago)

(5) all are divisible

RULE: add up all the digits in odd position of the number then add up all the digits in the even no.- If the difference of these two additions are 11 or 0 it means the no. is divisible by 11 else not.

(3+4+6+2)-(2+5+8)=0
(8+7+2)-(2+3+1)=11 (if the digits in even place is greater so we subtract the position of even place by position of odd place)
(4+8+1)-(3+4+6)=0
(3+8+4)-(6+9)=0

Is this solution Helpfull? Yes (1) | No   

Rubita  (9 years ago)

If you sum every second digit and then subtract all other digits and the answer is: 0, or, divisible by 11
Answer:
2)only B

Is this solution Helpfull? Yes | No (2)  

radha chowdaey  (9 years ago)

ALL ARE DIVISIBLE BY 11

Is this solution Helpfull? Yes | No (1)  

prashant kumar  (9 years ago)

all devisible by 11

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Prince Kumar Yadav  (9 years ago)

All are divisible. Option D.

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souharda  (9 years ago)

all r divisible
option 5
diff between sum of all odd and even places are zero or divisible by 11

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Dheeraj Kumar  (9 years ago)

1

sum of odd place digits-sum of even place digits=0

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pragya gupta  (9 years ago)

All are divisible.if sum of odd and even place digit is subtracted,and if it gives either 0 or 11.that is why all are divisible.

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SUNIL KUMAR YADAV  (9 years ago)

in given no 1st we have to check the difference of its( even place digit sum-odd place digit sum=0 0r 11then we can cay no is divisible by 11.

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sorav  (9 years ago)

5) is right answer because difference of sum of digits in odd places and sum of digits in even places is zero or divisible by 11

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Vivek Bhagwani  (9 years ago)

Answer is 5) All are divisible
divisibility by 11 is confirmed if (sum of odd digits - sum of even digits) is either 0 or multiple of 11.

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katasani Sankar Reddy  (9 years ago)

In the given options , Only B satisfies the condition...
283712 => (2+7+8) - (1+3+2) => 17-6 = 11
ans :B

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Santanu Pal  (9 years ago)

Answer is option 5)All are divisible
Rule for Divisibility By 11:If the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a number divisible by 11.
So, as per rule all the given numbers are divisible by 11

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DINESH KUMAR KK  (9 years ago)

divisibility test for 11 is (sum of even digit)-(sum of the odd digit)
equal to 11 or 0,here all value get 0 or 11
so ans is all are divisible

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nassarshahab  (9 years ago)

only b because eventerm-oddterm

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Ritu raj  (9 years ago)

5,all r divisible

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akshay sudhakar shedwad  (9 years ago)

all number are divisible by 11...check(sum of odd place)-(sum of even place)=ans
if ans is divisible by 11 then it divisible number and ans=0 then its true

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Kartal Singh Pramanik  (9 years ago)

Ans is 5) because the difference between the sum of its digits at odd places and the sum of its at even places is either 0 or a number divisible by 11
so all the given numbers are divisible by 11

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Neha V  (9 years ago)

option 5) is the right answer.
The divisiblility check for 11 is to subtract the sum of numbers obtained in odd and even places since all numbers satisy the criterion all are divisible.

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KANISHKA  (9 years ago)

x+x^2=1406
x^2+x-1406=0
by solving this quadratic eqn we get x=37

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sujatha c k  (9 years ago)

(sum of numbers at odd places)-(sum of numbers at even places) should be divisible by 11 so answer is (2+3+1)-(8+7+2)=11 which is divisible by 11
only b

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SUVATHINI.S  (9 years ago)

odd no-even no= 11 or 0 we get the answer

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sindu  (9 years ago)

2
(8+7+2)-(2+3+1)=11
11/11
it divisible by 11

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mahesh  (9 years ago)

ans:5;all are divisible

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RAJ  (9 years ago)

take the alternate number add them..we get two groups subtract that groups..if it is div then it will..

Is this solution Helpfull? Yes | No (1)  

Naveen  (9 years ago)

all are divisible.....check as if (sum of odd place digit-sum of even place digit)you will get either 0 or 11.hence all number is divisible by 11(if any daught then calculator pe kr k dekh lo)

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prakash  (9 years ago)

all are divisible

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surya  (9 years ago)

option is the correct

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Amit Kumar  (9 years ago)

ALL ARE DIVISIBLE SINCE WE GET THE DIFFERENCE OF SUM OF ODD PLACE AND EVEN PLACE AS 0(IN OPTION A,C,D) AND 11(IN OPTION B AS 2+7+8=17(ODD) AND 1+3+2=6(EVEN)
SO 17-6=11)

Is this solution Helpfull? Yes | No (1)  

B.Priyanka  (9 years ago)

All r divisible

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subhash  (9 years ago)

all are divisible by 11 so its 5

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seshadrivasam preethi  (8 years ago)

(2+7+8)-(1+3+2)=11

Is this solution Helpfull? Yes | No (1)  

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