Aptitude - Numbers - Discussion

Let the given number be, 476xy0.

It is given that 476xy0 is divisible by both 3 & 11.

To check the number is divisible by 3, we have to sum up all the digits, if the sum of the digits is divisible by 3 then the number is also divisible by 3.(a test of divisibility method).

From opt.A (x=7,y=4)
4+7+6+7+4+0=28 i.e, not divisible by 3.

From opt.B (x=7,y=5)
4+7+6+7+5+0=29 i.e, not divisible by 3.

From opt.C (x=8,y=5)
4+7+6+8+5+0=30 i.e, divisible by 3("").

Now,
Further, we have to check that opt.C i.e, x=8 & y=5 is eligible for the condition being divisible by 11 or not.

To check the number is divisible by 11,we have to find the difference of the sum of the digits at odd places & sum of the digits at even places,is either 0 or divisible by 11.(a test of divisibility method).

So, the number is 476850.
Sum of the odd places = 0+8+7=15,
Sum of the even places= 5+6+4=15,
Difference = (sum of the odd places)-(sum of the even places).
= 15-15.
= 0.
As the difference is '0' hence it is divisible by 11("").
Therefore,
The required numbers are,( x=8 & y=5).

How x=2 and x=8 came? please explain it.

First option - 7 and 4.
4 + 7 + 6 + 7 + 4 + 0 = 28/3 = 7.
(4 + 6 + 4 = 14) - (7+7+0= 14) = 0.

Thank you @Tanya.

The sum of the numbers should be divisible by 3 in order for the number to be completely divisible by 3.

So 4+7+6+0=17.

Adding the numbers in the options so that the total number is divisible by 3.

17+ 8+5 = 30.
Hence option C is correct.

Hi, it is very simple.

4+7+6+8+5=30 is completely divisible by 3.
And according to the law of algebra < (5+6+4 )-(8+7)=15-15=0 it is divisible by 11.
So, The non-zero digits in the hundred's and ten's places are respectively: 8 and 5.

Let the given number be 476 xy 0.

add the digits
(4 + 7 + 6 + x + y + 0) = (17 + x + y) must be divisible by 3.

To check no. is divisible by 3, the easiest way is to check from the options:

from A. put 17+7+4=17+11=28 not divisible by 3
from B. put 17+7+5=17+12=29 not divisible by 3
from C. put 17+8+5=17+13=30 is divisible by 3

therefore,x=8 and y=5.

To check no. is divisible by 11:
(0 + x + 7) - (y + 6 + 4) = (x - y -3) must be either 0 or 11.
from A. put 7-4-3=7-7=0 is divisible by 11
from B. put 7-5-3=7-8=-1 not divisible by 11
from C. put 8-5-3=8-8=0 is divisible by 11

therefore,x=8 and y=5.

Option C is correct.

2x+14 what about this term?

Can someone please explain and x=2 and x=8 from where it is overcome please explain?

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?

I can't understand please explain me from first step to last step.