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# After multiplying a positive integer A, which has n digits,

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After multiplying a positive integer A, which has n digits,  [#permalink]

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15 Feb 2013, 11:04
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Question Stats:

34% (01:30) correct 66% (01:07) wrong based on 435 sessions

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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

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Re: After multiplying a positive integer A, which has n digits,  [#permalink]

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15 Feb 2013, 18:05
21
3
emmak wrote:
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

The question seems convoluted but it's not. You have to take the first step in the right direction. The only definitive thing given here is that we get a number with (n+1) digits, all the digits being (n+1). What will such a number look like?

22
333
4444
55555 etc

We obtain this number by multiplying A with (n+2). This means that our number should be divisible by (n+2). Now, ask yourself:
Is 22 divisible by 3? No.
Is 333 divisible by 4? No
We know that no odd number will be divisible by an even number. So we can ignore 333, 55555, 7777777 etc

Only consider even numbers:

Is 4444 divisible by 5? No

Is 666666 divisible by 7? Yes! Check: 666666/7 = 95238 (5 digit number). SO when you multiply 95238 by 7, you get 666666

Is 88888888 divisible by 9? No

Use divisibility rules to quickly rule out the numbers not divisible.

_________________

Karishma
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##### General Discussion
MBA Section Director
Affiliations: GMAT Club
Joined: 21 Feb 2012
Posts: 5991
City: Pune
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

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15 Feb 2013, 11:56
emmak wrote:
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?

None

1

2

8

9

Constraint 1) when we put a sequence of multiples of (n+2) atleast one multiple should have its unit digit same as that of (n+1)
Constraint 2) N+1 can not greater than 9 since it is a single digit.
Constraint 2) N can not be 0

n(n+1)(n+2)
8---9----10------ Units digit zero always. so out
7---8----9-------- 9X2=18 so 2323232X9 = .......2988 or 2222222 x 9 = ......98 out
6---7----8-------- units digits 8,6,4,2,4,8,6,4,2,0,8,6.... No 7 so out
5---6----7-------- 7x8=56 so 83838 x 7 = .......5866 or 88888 x 7 = ....216 out
4---5----6-------- units digits 6,2,8,4,0,6,2,8,4,0...... No 5 so out
3---4----5-------- 5,0,5,0 out
2---3----4-------- 4,8,2,6,0,4.... out
1---2----3-------- only possible pair is 3 x 4 = 12 so out

Bunuel, Can you Pls help?
_________________
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Re: After multiplying a positive integer A, which has n digits,  [#permalink]

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16 Feb 2013, 08:14
VeritasPrepKarishma wrote:
emmak wrote:
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

The question seems convoluted but it's not. You have to take the first step in the right direction. The only definitive thing given here is that we get a number with (n+1) digits, all the digits being (n+1). What will such a number look like?

22
333
4444
55555 etc

We obtain this number by multiplying A with (n+2). This means that our number should be divisible by (n+2). Now, ask yourself:
Is 22 divisible by 3? No.
Is 333 divisible by 4? No
We know that no odd number will be divisible by an even number. So we can ignore 333, 55555, 7777777 etc

Only consider even numbers:

Is 4444 divisible by 5? No

Is 666666 divisible by 7? Yes! Check: 666666/7 = 95238 (5 digit number). SO when you multiply 95238 by 7, you get 666666

Is 88888888 divisible by 9? No

Use divisibility rules to quickly rule out the numbers not divisible.

You said correctly Karishma.
It is important to take first step in right direction.

Regards,

Abhijit
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Re: After multiplying a positive integer A, which has n digits,  [#permalink]

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19 Aug 2018, 12:26
emmak wrote:
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

The question seems convoluted but it's not. You have to take the first step in the right direction. The only definitive thing given here is that we get a number with (n+1) digits, all the digits being (n+1). What will such a number look like?

22
333
4444
55555 etc

We obtain this number by multiplying A with (n+2). This means that our number should be divisible by (n+2). Now, ask yourself:
Is 22 divisible by 3? No.
Is 333 divisible by 4? No
We know that no odd number will be divisible by an even number. So we can ignore 333, 55555, 7777777 etc

Only consider even numbers:

Is 4444 divisible by 5? No

Is 666666 divisible by 7? Yes! Check: 666666/7 = 95238 (5 digit number). SO when you multiply 95238 by 7, you get 666666

Is 88888888 divisible by 9? No

Use divisibility rules to quickly rule out the numbers not divisible.

Hi Karishma,

How did you reach to 666666 specifically?

Also, the qualifying numbers in the problem could be
11,22,33,44,55,66,77,88,99
111,222,333,444,555,666,777,888,999
1111....
11111....
111111...
and on

How do we narrow down this in a short span of time? Sorry, I could not understand this.

Thanks,
Vikram
Director
Joined: 20 Sep 2016
Posts: 555
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

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22 Aug 2018, 23:31
1
vikgupta07 wrote:
emmak wrote:
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

The question seems convoluted but it's not. You have to take the first step in the right direction. The only definitive thing given here is that we get a number with (n+1) digits, all the digits being (n+1). What will such a number look like?

22
333
4444
55555 etc

We obtain this number by multiplying A with (n+2). This means that our number should be divisible by (n+2). Now, ask yourself:
Is 22 divisible by 3? No.
Is 333 divisible by 4? No
We know that no odd number will be divisible by an even number. So we can ignore 333, 55555, 7777777 etc

Only consider even numbers:

Is 4444 divisible by 5? No

Is 666666 divisible by 7? Yes! Check: 666666/7 = 95238 (5 digit number). SO when you multiply 95238 by 7, you get 666666

Is 88888888 divisible by 9? No

Use divisibility rules to quickly rule out the numbers not divisible.

Hi Karishma,

How did you reach to 666666 specifically?

Also, the qualifying numbers in the problem could be
11,22,33,44,55,66,77,88,99
111,222,333,444,555,666,777,888,999
1111....
11111....
111111...
and on

How do we narrow down this in a short span of time? Sorry, I could not understand this.

Thanks,
Vikram

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?

You say 11 could be one of the numbers- here's how it cannot-

11 is the product of A with n number of digits and (n+2).
Now to get 11 we have to have number of digits to be 0. Only then can the number be (n+1=0+1) and as the question tells us that all the numbers are all same then in no way can you get 11 a 2 digit number of n =0 as the maximum the product could be is of 1 digit.

Posted from my mobile device
Intern
Joined: 30 May 2017
Posts: 13
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

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23 Aug 2018, 02:53
Hello,
I do not understand the question and the answers.
I cannot see written in the text that all digits of A are the same.

let's say that A = 4,5,6,7, 8 or 9 in this case n=1. If I multiply A by 3 (n+2) I will have a number with 2 digits.

If I choose a number with two digits: A= 25, 26, 27,..., 99 then A*4 will have 3 digits and so on..

So what didn't I understand?
Director
Joined: 20 Sep 2016
Posts: 555
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

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23 Aug 2018, 03:07
amascarenhas wrote:
Hello,
I do not understand the question and the answers.
I cannot see written in the text that all digits of A are the same.

let's say that A = 4,5,6,7, 8 or 9 in this case n=1. If I multiply A by 3 (n+2) I will have a number with 2 digits.

If I choose a number with two digits: A= 25, 26, 27,..., 99 then A*4 will have 3 digits and so on..

So what didn't I understand?

You may have been right about not understanding the question.

You got the first 2 parts right.

Third part- we get a number with (n+1) digits, all of whose digits are (n+1).

This means that product of A and n+2 has every digit same.
As in A= 95238 >>number of digits (n)= 5

Therefore A*(5+2) = 666666

The answer has 6 ie.(n+1) digits . And each digit in the answer is of the form (n+1)

So basically if you consider
Say A=23 >> n=2
A* (n+2) = (n+1)*(n+1)*(n+1)

Here the answer has three digits ie. (n+1) digit with each digit as (n+1)

Now the above example actually does not exist as there is only one such number and that is 666666

Posted from my mobile device
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8789
Location: Pune, India
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

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23 Aug 2018, 07:49
amascarenhas wrote:
Hello,
I do not understand the question and the answers.
I cannot see written in the text that all digits of A are the same.

let's say that A = 4,5,6,7, 8 or 9 in this case n=1. If I multiply A by 3 (n+2) I will have a number with 2 digits.

If I choose a number with two digits: A= 25, 26, 27,..., 99 then A*4 will have 3 digits and so on..

So what didn't I understand?

Read this: "... we get a number with (n+1) digits, all of whose digits are (n+1)"

The number has (n+1) digits. Each digit is (n+1).
So the possible numbers are:
1 digit number with each digit 1 i.e. 1
2 digit number with each digit 2 i.e. 22
3 digit number with each digit 3 i.e. 333
and so on...
_________________

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Veritas Prep GMAT Instructor

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Re: After multiplying a positive integer A, which has n digits,  [#permalink]

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24 Aug 2018, 10:07
emmak wrote:
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

_________________

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Intern
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Posts: 6
Location: India
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

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27 Aug 2018, 01:56
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?

You say 11 could be one of the numbers- here's how it cannot-

11 is the product of A with n number of digits and (n+2).
Now to get 11 we have to have number of digits to be 0. Only then can the number be (n+1=0+1) and as the question tells us that all the numbers are all same then in no way can you get 11 a 2 digit number of n =0 as the maximum the product could be is of 1 digit.

Posted from my mobile device[/quote]

Thank you!

Your comment did somehow helped me solve my doubt. I got it sorted out.
Intern
Joined: 17 Aug 2015
Posts: 6
Location: India
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

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27 Aug 2018, 02:04
780gmatpossible wrote:
emmak wrote:
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

Hi,

Let me try and explain

1. A=Positive Integer [N Digits], Let's say 5
2. Multiplication Factor: [N+2], From 1, it's 5+2=7
3. Resultant Number: 5*7=35, which has [N+1] digits, according to the question, but all digits are not [N+1], The required number, according to this case should have been 66

The only numbers that would satisfy the problem are: -

1
22
333
4444
55555
666666
7777777

and so on.

If this is the resultant number, then if it has [N] digits, it must have been divisible by [N+1], FROM 2 Above. Do the quick math as to what number is divisible by [N+1]

from the qualifying set above, 666666/7=95238 is a candidate.

This A should satisfy all boundaries of the problem. therefore instances=1 [B ]

Hope this helps.

Thanks,
VIKRAM
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 616
Re: After multiplying a positive integer A, which has n digits,  [#permalink]

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17 Dec 2018, 04:11
emmak wrote:
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

$$\left. \matrix{ A = \,\,\underbrace {\underline {} \,\,\underline {} \,\,\underline {} \,\, \ldots \,\,\underline {} }_{n\,\,{\rm{digits}}}\,\,\,\, \ge 1\,\,{\mathop{\rm int}} \hfill \cr A \cdot \left( {n + 2} \right) = \underbrace {\underline {n + 1} \,\,\underline {n + 1} \,\,\underline {n + 1} \,\, \ldots \,\,\underline {n + 1} }_{n + 1\,\,{\rm{digits}}}\,\,\,\, \hfill \cr} \right\}\,\,\,\,\,\,\,\,?\,\,\,\, = \,\,\,\,\# \,\,A\,\,\,{\rm{possible}}$$

This is a typical organized manual work technique exercise!

$$\left\{ \matrix{ n = 1\,\,\,\,\, \Rightarrow \,\,\,\,\underline {} \,\, \cdot \,\,\,\left( {1 + 2} \right) = \underline 2 \,\,\underline 2 \,\,\,\,,\,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{{22} \over 3} \ne {\mathop{\rm int}} } \right) \hfill \cr n = 2\,\,\,\,\, \Rightarrow \,\,\,\,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {2 + 2} \right) = \underline 3 \,\,\underline 3 \,\,\underline 3 \,\,\,\,\,,\,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{{{\rm{odd}}} \over {{\rm{even}}}} \ne {\mathop{\rm int}} } \right) \hfill \cr n = 3\,\,\,\,\, \Rightarrow \,\,\,\,\underline {} \,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {3 + 2} \right) = \underline 4 \,\,\underline 4 \,\,\underline 4 \,\,\underline 4 \,\,\,\,\,\,\,,\,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{{4444} \over 5} \ne {\mathop{\rm int}} } \right) \hfill \cr n = 4\,\,\,\,\, \Rightarrow \,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{\rm{idem}}\,\,n = 2} \right) \hfill \cr n = 5\,\,\,\,\, \Rightarrow \,\,\,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {5 + 2} \right) = \underline 6 \,\,\underline 6 \,\,\underline 6 \,\,\underline 6 \,\,\underline 6 \,\,\underline 6 \,\,\,\,\,\,\,,\,\,\,\,\underline {{\rm{viable}}} \,\,{\rm{solution}}\,\,\,\,\,\left( {{{666666} \over 7} = 95238} \right) \hfill \cr n = 6\,\,\,\,\, \Rightarrow \,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{\rm{idem}}\,\,n = 2} \right) \hfill \cr n = 7\,\,\,\,\, \Rightarrow \,\,\,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {7 + 2} \right) = \underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\underline 8 \,\,\,\,\,,\,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{{88888888} \over 9} \ne {\mathop{\rm int}} } \right) \hfill \cr n = 8\,\,\,\,\, \Rightarrow \,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{\rm{idem}}\,\,n = 2} \right) \hfill \cr n = 9\,\,\,\,\, \Rightarrow \,\,\,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\underline {} \,\, \cdot \,\,\,\left( {9 + 2} \right) = \underline {10} \,\,\underline {10} \,\,\underline {10} \,\, \ldots \,\,\underline {10} \,\,\,????\,\,\,\,{\rm{impossible}}\, \hfill \cr n \ge 10\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{impossible}}\,\,\,\,\,\left( {{\rm{idem}}\,\,n = 9} \right) \hfill \cr} \right.$$

The correct answer is therefore (B). (This is all VERY fast, although hard to type!)

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: After multiplying a positive integer A, which has n digits, &nbs [#permalink] 17 Dec 2018, 04:11
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