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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,

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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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alanforde800Maximus 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.

_________________
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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alanforde800Maximus 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
Let’s test some values for n.

If n = 1, A is a 1-digit number. We multiply A by 3 and we get a 2-digit number, which is 22. However, 22 is not a multiple of 3. So n can’t be 1.

If n = 2, A is a 2-digit number. We multiply A by 4 and we get a 3-digit number, which is 333. However, 333 is not a multiple of 4. So n can’t be 2.

If n = 3, A is a 3-digit number. We multiply A by 5 and we get a 4-digit number, which is 4444. However, 4444 is not a multiple of 5. So n can’t be 3.

If n = 4, A is a 4-digit number. We multiply A by 6 and we get a 5-digit number, which is 55,555. However, 55,555 is not a multiple of 6. So n can’t be 4.

If n = 5, A is a 5-digit number. We multiply A by 7 and we get a 6-digit number, which is 666,666. We see that 666,666 is a multiple of 7 (666,666 = 7 x 95,238)! So n can be 5.

At this point, we can skip even values of n, since the (n+1)-digit number it forms is odd and will never be a number of n + 2, which is even.

If n = 7, A is a 7-digit number. We multiply A by 9 and we get a 6-digit number, which is 88,888,888. However, 88,888,888 is not a multiple of 9. So n can’t be 7.

If n = 9, A is a 9-digit number. We multiply A by 11 and we get a 10-digit number. However, we can’t have a 10-digit number in which each of its digits is 10. Therefore, n can’t be 9, and we can stop here.

There is only one instance, n = 5, where all the criteria are met.

Answer: B

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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

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

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Contact Rich at Rich.C@empowergmat.com

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