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After multiplying a positive integer A, which has n digits,
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15 Feb 2013, 11:04
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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,
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15 Feb 2013, 18:05
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,
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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) 8910 Units digit zero always. so out 789 9X2=18 so 2323232X9 = .......2988 or 2222222 x 9 = ......98 out 678 units digits 8,6,4,2,4,8,6,4,2,0,8,6.... No 7 so out 567 7x8=56 so 83838 x 7 = .......5866 or 88888 x 7 = ....216 out 456 units digits 6,2,8,4,0,6,2,8,4,0...... No 5 so out 345 5,0,5,0 out 234 4,8,2,6,0,4.... out 123 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,
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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. 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,
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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,
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22 Aug 2018, 23:31
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. Posted from my mobile device



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Re: After multiplying a positive integer A, which has n digits,
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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?



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Re: After multiplying a positive integer A, which has n digits,
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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



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Re: After multiplying a positive integer A, which has n digits,
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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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Re: After multiplying a positive integer A, which has n digits,
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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 Bunuel  Can you please help with this one. Thanks
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Re: After multiplying a positive integer A, which has n digits,
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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?
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,
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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 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,
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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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