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.
Akash said:
1 decade ago
(4 + 7 + 6 + x + y + 0) = (17 + x + y) must be divisible by 3.
How it can be said it must be divisible by 3. Please explain?
How it can be said it must be divisible by 3. Please explain?
Sudhir said:
1 decade ago
Actually rule for divisbility of 11 is sum of odd places - even places but you did even -odd in the problem. Is it 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?
Sumit jain said:
1 decade ago
The 1st option is also suiting well in the sum. Then the numbers ll be divisible by both 11 and 3 ?
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.
Vishnu Priya said:
9 years ago
Please can anyone explain from (0 + x + 7) - (y + 6 + 4) =x - y - 3. Must be either 0 or 11?
Rahul said:
9 years ago
By solving through divisibility method we can aslo get value X = 5 and Y = 2. Is it correct?
Dhanashri said:
9 years ago
Please explain this step, (0 + x + 7) - (y + 6 + 4) = (x - y -3) must be either 0 or 11.
Roxy said:
1 decade ago
Just try hit and trial method to solve it, by putting the values in the missing places.
Durga prasad said:
1 decade ago
In explanation, I can't understand last 2 steps.
Please anybody explain them.
Please anybody explain them.
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