Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 51)
51.
476 ** 0 is divisible by both 3 and 11. The non-zero digits in the hundred's and ten's places are respectively:
Answer: Option
Explanation:
Let the given number be 476 xy 0.
Then (4 + 7 + 6 + x + y + 0) = (17 + x + y) must be divisible by 3.
And, (0 + x + 7) - (y + 6 + 4) = (x - y -3) must be either 0 or 11.
x - y - 3 = 0 
y = x - 3
(17 + x + y) = (17 + x + x - 3) = (2x + 14)
x= 2 or x = 8.
x = 8 and y = 5.
Discussion:
37 comments Page 2 of 4.
Shahid ali said:
8 years ago
Thank you @Tanya.
Pritam Bose said:
8 years ago
Thank you @Tanya Mazumdar.
Rinkal said:
8 years ago
Thanks @Tanya Mazumdar.
Jashim uddin said:
7 years ago
Thanks @Tanya.
Shekhar said:
7 years ago
How can come x=2 or 8?
Please explain.
Please explain.
Raj said:
7 years ago
Divisibility by 3 condition is sum of all the digits are divisible by 3.
Divisibility by 11 condition is diff of sum of odd and even place digits should be equal to zero.
Let the missing digits be x and y.
Then given value becomes 476xy0.
For divisible by 11: (4+6+y)-(7+x+0)=0
=> 10+y=7+x.
=> y = (7-10)+x.
=> y = x-3 ------> (1).
Now for divisibility by 3: 4+6+7+x+y+0 => 17+x+y.
=> 17+x+x-3 => 14+2x should be divisible by 3,
now try substuiting values:
x=1: 16 not div by 3,
x=2: 18 div by 3,
x=3: 20 not div by 3.
.
.
x=5: 24 div by 3.
So for x=2
=> y= -1 which is not valid.
=> for x= 5
=> y=2. then 10+y =7+x
so the answer 2 and 5.
It will work for 8 and 5.
Divisibility by 11 condition is diff of sum of odd and even place digits should be equal to zero.
Let the missing digits be x and y.
Then given value becomes 476xy0.
For divisible by 11: (4+6+y)-(7+x+0)=0
=> 10+y=7+x.
=> y = (7-10)+x.
=> y = x-3 ------> (1).
Now for divisibility by 3: 4+6+7+x+y+0 => 17+x+y.
=> 17+x+x-3 => 14+2x should be divisible by 3,
now try substuiting values:
x=1: 16 not div by 3,
x=2: 18 div by 3,
x=3: 20 not div by 3.
.
.
x=5: 24 div by 3.
So for x=2
=> y= -1 which is not valid.
=> for x= 5
=> y=2. then 10+y =7+x
so the answer 2 and 5.
It will work for 8 and 5.
Govind giri said:
7 years ago
For checking the divisibility by 11 we look from the most significant digit for even and odd places but in this question the process is reversed. Is this correct?
Ravina said:
6 years ago
2x+14 what about this term?
Can someone please explain and x=2 and x=8 from where it is overcome please explain?
Can someone please explain and x=2 and x=8 from where it is overcome please explain?
Pranab said:
8 months ago
First option - 7 and 4.
4 + 7 + 6 + 7 + 4 + 0 = 28/3 = 7.
(4 + 6 + 4 = 14) - (7+7+0= 14) = 0.
4 + 7 + 6 + 7 + 4 + 0 = 28/3 = 7.
(4 + 6 + 4 = 14) - (7+7+0= 14) = 0.
Roxy said:
1 decade ago
Just try hit and trial method to solve it, by putting the values in the missing places.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers