Aptitude - Numbers - Discussion

The divisibility rule of 3 is to add all the digits, that number should be divisible by 3.

Here 5+1+7+*+3+2+4 = 22+*.

If we add 2 to it, then22+2 = 24.
Where 24 is divisible by 3.
Therefore, the value of * is 2.

As per the Divisibility Rule - A number is divisible by 3, if the sum of all its digits is also divisible by 3.

@Suresh @Mariya

Yeah, let's assume if
SAM is divisible by 3
S+A+M is also divisible by 3.

5 + 1 + 7 + x + 3 + 2 + 4 if we add this it would be 22+x.

So 22 is not divisible by 3.
If we add 2 in place of x i.e, 22+2 = 24 then it is divisible by 3.
So the answer is "2".

Thank you, everyone, l have understood the answer.

According to you guys if a number is divisible by another number then sum of its all digits also divisible. But it's not true in all cases.

e.g. The number is 625 which is divisible by 5. But if we add all its digits 6+2+5=13 which is not divisible by 5.

So this principle is wrong.

@Aravind
456*85.
1. 4+5+6+x+8+5=28+x
2. 28+10(A is given option )=38/3=12.66 ஃ A is No
3. 28+84=112/3=37.333. No
4. 28+12=/3=13.33. No
5.28+21=16.33. No

Ans: None of those

But.
28+2=30/3=10. Ans is 2.

The answer is 1.

Hi paavani..

Principle: If a number is divisible by 3 means, sum of the digits also divisible by 3.

For eg: take 12. we know that 12 is divisible by 3
sum of digits i.e 1+2 = 3 ,3 is divisible by 3.

First u clear about this?

We cum to problem then..
517*324 is completely divisible by 3
so sum of the digits also divisible by 3.

Then, 5+1+7+x+3+2+4 = 22+x

We hav given the options 0, 1 , 2

If x=0, sum= x+22= 0+22 = 22 is not divisible by 3
If x=1, Sum =x+22 = 1+22 = 23 is not divisible by 3
If x=2, Sum =x+22 = 2+22 = 24 is divisible by 3
So answer is 2

Hope you understand this..

I have simply used the given options to solve the question.

517*324. Replace the * by 0 and divide by 3.

The answer doesn't come in the whole number which means it's not divisible. But when you replace the * by 2 the answer comes in the whole number.

Thanks everyone.