Aptitude - Numbers - Discussion

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.

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

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).

2x+14 what about this term?

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

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.

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?

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.

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.

How can come x=2 or 8?

Please explain.