In the above number, A and B represent the tens and units digits

Question

110AB
In the above number, A and B represent the tens and units digits, respectively. If the above number is divisible by 55, what is the greatest possible value of B x A?

A. 0
B. 5
C. 10
D. 15
E. 25

camlan1990 · Answer

Two difit number, which is divisible by 55, can be 11 or 55

The largest value of B X A =5*5 = 25

Ans: E  Please corrent if I'm wrong. Thanks

EMPOWERgmatRichC · Answer

Hi Bunuel,

The prompt refers to an 'above number', but does not include a number for reference. Is this prompt complete?

GMAT assassins aren't born, they're made,
Rich

Amidro · Answer

I also was confused when I was looking for "above number" :D

As far as I understood, 55 is a factor of AB. In other words, the values of B (units digits can be 5 or 0. Better to have option for 5 in this case to have "bigger result"). Now let's try 55x1 (A & B=5 respectively we have =25). To ensure, let's check (avoid even multiples of 2,4,6,8 etc (we will have 0 in units thus making our result 0)
55x3 = 165 (A=6 B=5 respectively. Hey! that's 30. But we don't have this number in answer choices, move on).
55x5= 275 (A=7, B=5 respectively. Increasing trend , we have 35 now)
55x7= 385 (A=8, B=5. Have 40 now. can create a pattern)
IMO E.
But here we can have different answer choices and I think that :
1. I have misunderstood task.
2. The question is not full.

Anyway you can KUDOS me if you find anything useful in my post

Bunuel · Answer

Sorry! Edited the question. Please try again. Thank you,

Abhishek009 · Answer

\frac{110AB}{55} = Remainder 0

\frac{11000}{55}   {Minimum} = Remainder 0  { \frac{11000}{55}  = 200}

Now  \frac{AB}{55}  = Must have Remainder 0 and it must be a 2 digit number.

So the only value ( Maximum as well as minimum ) of AB = 55

Thus the number becomes -

\frac{11055}{55} = 201

A*B = 5 * 5 => 25

HarrishGowtham · Answer

This can be done in 2 ways.

1(plug and play ).

Given that 110AB is divisible by 55 (note that 110 is in itself divisible by 55). So the other 2 digits can possibly be either 00 or 55. Since we look for the greatest positive integer, AB should be 55 making the product 25.

2nd method. 55 is a multiple of 11 and hence the divisibility for 11 is also valid here. (sum of digits in odd place-sum of digits in even should be either 0 or 11).
i.e. (1+0+B)-(1+A)=0 OR 11
 implies B-A=0
The only possible values being 00. or 55.

Answer choice E.

stonecold · Answer

Here is what i did =>
If a number is divisible by 55 => it must be divisible by 5 as well as 11
As it is divisible by 5 => unit digit can be 0 or 5 
If the unit digit is 0 => Product of A*B=0 irrespective of what A will be.
lets check for when B=5
Hmm then for the number to be divisible by 11 => A must be 5 too
Hence AB=25
And we can stop playing with the digits as the highest value in the options is 25

Hence E

Lets se

ScottTargetTestPrep · Answer

Since we are given that 110AB/55 = integer, let’s start by determining the possible values of 110AB. Since, 110 is divisible by 55, 11,000 must also be divisible by 55. Thus, the next highest integer that is divisible by 55 is 11,055. Since 11,000 and 11,055 are the only two values that can equal 110AB, we see that the maximum value of B x A is 5 x 5 = 25 (since the other value is 0 x 0 = 0).

Answer: E

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In the above number, A and B represent the tens and units digits

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In the above number, A and B represent the tens and units digits

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