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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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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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New post 15 Feb 2013, 18:05
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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.

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

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New post 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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New post 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.

Answer (B)


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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New post 19 Aug 2018, 12:26
VeritasKarishma 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.

Answer (B)



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

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New post 22 Aug 2018, 23:31
1
vikgupta07 wrote:
VeritasKarishma 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.

Answer (B)



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?


Read the question carefully:

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.

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

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

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New post 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

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

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

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New post 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


Bunuel - Can you please help with this one. Thanks
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Re: After multiplying a positive integer A, which has n digits,  [#permalink]

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New post 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?


Read the question carefully:

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

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New post 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


Bunuel - Can you please help with this one. Thanks


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

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New post 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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