Aptitude - Numbers - Discussion

Discussion Forum : Numbers - General Questions (Q.No. 8)
8.
The largest 4 digit number exactly divisible by 88 is:
9944
9768
9988
8888
None of these
Answer: Option
Explanation:
Largest 4-digit number = 9999

 88) 9999 (113
     88
     ----
     119
      88
     ----
      319
      264
      ---
       55
      ---
       
 Required number = (9999 - 55)
                 = 9944.        
Discussion:
67 comments Page 2 of 7.

Darshan Prajapati said:   4 years ago
How it is correct? because above in the example no. 6 method is to split the number & then check the number either diveded by all or not.

And same method if we applied in this example (11*2*4) = 88.

So, 9988 is divided by all 11, 2, & 4.

Please anyone clarify this.
(3)

Murali goud said:   7 years ago
We can write 88=11*8=11*4*2.

So the 4 digit number must be divisible by all the digits i.e 11,4,2.

But where am I confusing is the number 9988 is divisible by all the numbers i.e 11,4,2 but it is not divisible by 88.

Why? The method is wrong or write? Can anyone suggest?
(2)

Nantha said:   2 years ago
@Darshan Prajapati.

We should treat the number as 2×4 and not 2, 4.88=11×2^3.
So, it should be divisible by 8 and not just by 4.
(2)

Aditya said:   6 years ago
The easiest way to understand this is Euclid's division lemma.

Dividend = divisior * quatient+remainder.
So 9999 = 88 * 113 + 55..
9999-55 = 88 * 113 = 9944 = 9944.
(1)

Pankaj Namekar said:   7 years ago
9988 is the answer because;
88= (11*2*4).
9988 is completely divisible by 88.
(1)

Aditya said:   8 years ago
Is it possible answer is 9876?

Gitanjali said:   8 years ago
What if the question is smallest 4 digit number. I know we need to divide with 1000 but what we'll do after divisions like here it is 9999 - 55, what will be there?

Sushma said:   8 years ago
But the divisibility rule of 8 is last 3 digits should be divisible by 8 not the first 3 digits.

Sowmiya said:   8 years ago
Why we take (9999-55)?

Ade said:   8 years ago
How did you get 264?


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