Aptitude - Numbers - Discussion

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?

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.

2x+14 what about this term?

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

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.

Thank you @Tanya.

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