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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] New post 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
[Reveal] Spoiler: OA

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Re: After multiplying a positive integer A, which has n digits, [#permalink] 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] New post 15 Feb 2013, 11:56
Expert's post
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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Expert Post
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Affiliations: GMAT Club
Joined: 21 Feb 2012
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Re: After multiplying a positive integer A, which has n digits, [#permalink] New post 16 Feb 2013, 08:14
Expert's post
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] 16 Feb 2013, 08:14
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