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After multiplying a positive integer A, which has

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by Gmat_mission » Fri Jun 01, 2018 1:27 am

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After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

A. None
B. 1
C. 2
D. 8
E. 9

[spoiler]OA=B[/spoiler]

I don't have any idea of how to solve this PS question. <i class="em em-sob"></i> Could someone give me an explanation? Please. Thanks in advance.
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Source: — Problem Solving |

by [email protected] » Fri Jun 01, 2018 9:39 am
Hi Gmat_mission,

We're told that after multiplying a positive integer A (which has N digits) by (N+2), we get a number with (N+1) digits, all of whose DIGITS are (N+1). We're asked for the number of possible values of A that 'fit' this description.

To start, this question certainly 'feels' weird - and you would likely find it easiest to 'work back' from the later pieces of information that you're given - and use 'brute force' (along with some Number Properties) to find the solution. We're looking to end up with a number that has (N+1) digits - and ALL of those DIGITS equal (N+1). Since we're dealing with digits, there are only a limited number of possible values that we can end up with:

22
333
4444
55555
666666
7777777
88888888
999999999

Thus, there are no more than 8 possibilities; to get the correct answer, we have to incorporate the other pieces of information that we're given and see which of these end numbers actually 'fits' everything that we're told.
IF....
The end result was 22, then N=1, but there is no 1-digit number that you can multiply by (1+2) = 3 and end up with 22. This is NOT possible.
The end result was 333, then N=2, but there is no 2-digit number that you can multiply by (2+2) = 4 and end up an ODD. This is NOT possible.
The end result was 4444, then N=3, but there is no 3-digit number that you can multiply by (3+2) = 5 and end up with 4444. This is NOT possible.
The end result was 55555, then N=4, but there is no 4-digit number that you can multiply by (4+2) = 6 and end up with ODD. This is NOT possible.

The end result was 666666, then N=5... there IS a 5-digit number that you can multiply by (5+2) = 7 and end up with 666666 (it's 95,238 - you just have to do a little division to prove it). This IS a possibility.

The end result was 7777777, then N=6, but there is no 6-digit number that you can multiply by (6+2) = 8 and end up with ODD. This is NOT possible.
The end result was 88888888, then N=7, but there is no 7-digit number that you can multiply by (7+2) = 9 (since 88888888 is NOT a multiple of 9). This is NOT possible.
The end result was 999999999, then N=8, but there is no 8-digit number that you can multiply by (8+2) = 10 and end up with ODD. This is NOT possible.

Thus, there's just one answer that 'fits' everything that we're told.

Final Answer: B

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by deloitte247 » Sat Jun 02, 2018 2:04 pm
What have a number a with (n+1) digits the number there will look like
2 2 = [A*(n+2)]
3 3 3 =[A*(n+2)]
4 4 4 4 =[A*(n+2)]
5 5 5 5 5 =[A*(n+2)]
6 6 6 6 6 6 = ; and so on e. t. c. These figures are obtained by multiplying A with (n+2). Therefore our number should be divisible by (n+2) knowing that no odd number will be divisible by an even number, we will have to deal withe numbers alone.
22/3 =7.3 {not divisible}
4444/5 =888.5 {not divisible}
666666/7 =9528 {This is DIVISIBLE}
88888888/9 =9876543.111 {not divisible} using divisibility rules, we will be able to rule out indivisible numbers.
My answer is Option B as i have 1 instance with 666666
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