Consecutive integers problem

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by sushantgupta » Sun Jul 03, 2011 9:53 pm
every place can possibly have 9 digits. SO the answer is 9^5

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by amit2k9 » Sun Jul 03, 2011 10:47 pm
the number can be 12121
for 5th digit from right to left - 9
for 4th digit - 9 (zero comes into picture)
for 3rd digit - 9 (digit for 4th place neglected)
similarly for 2nd and 1st position 9 each.

9^5.
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by prashant misra » Mon Nov 21, 2011 7:34 am
the answer to this question is option A for sure we can set the anagram frid for 1st digit we have 9 choices excluding 0 2nd digit 9 choices including zero,3rd digit 8 choices,4th digit 7 choices and last digit 6 choices.

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by ronnie1985 » Thu Mar 29, 2012 7:54 am
(A) is answer
QED
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by him1985 » Thu Apr 12, 2012 12:08 am
ronnie1985 wrote:(A) is answer
QED
Hi ronnie,
correct answer is C.
Read the question carefully.
It says, no two CONSECUTIVE numbers are same,
you have considered no two numbers are same.......

:)
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by Lifetron » Wed Aug 29, 2012 7:19 am
9*9*9*9*9 = 9^5

First digit can't be '0' as it is a 5-digit number. But others can have '0' as a digit hence 9 possibilities.

Answer is C

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by chris558 » Mon Sep 24, 2012 7:32 am
C) 9^5

I used the slot method...

First slot can be any digit but 0. Therefore 10-1=0
Second slot can be any digit but whatever was in the first slot, including 0. Therefore 10-1=9
Third slow can be whatever digit except preceding one... 9
Etc...

9*9*9*9*9=9^5

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by rajeshsinghgmat » Fri Apr 05, 2013 2:07 am
(A) 9*9*8*7*6

Left, First digit cannot be 0.

Therefore, Left, First digit has 9 choices (1, 2, 3, 4, 5, 6, 7, 8, 9).

Now, Left, Second digit has 9 choices as one of the 10 digits ((0),1,2,3,4,5,6,7,8,9) has been used.

Now, Left, Third digit has 8 choices as two of the 10 digits (0,1,2,3,4,5,6,7,8,9) has been used.

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by Brent@GMATPrepNow » Fri Apr 05, 2013 5:50 am
chaitanyareddy wrote:How many 5-digit positive integers exist where no two consecutive digits are the
same?
A.) 9*9*8*7*6
B.) 9*9*8*8*8
C.) 9^5
D.) 9*8^4
E.) 10*9^4
Here's a step by step solution using the Fundamental Counting Principle (aka "slot method")

Take the task of building a 5-digit number and break it into stages.

Stage 1: Select the 1st digit (the ten thousands digit)
The first digit can be 1,2,3,4,5,6,7,8 or 9 (can't have 0 as first digit, otherwise it's not a 5-digit number)
So, we can accomplish stage 1 in 9 ways.

Stage 2: Select the 2nd digit (the thousands digit)
Once the 1st digit has been selected, we cannot select it for the 2nd digit
So, we can accomplish stage 2 in 9 ways.

Stage 3: Select the 3rd digit (the hundreds digit)
Once the 2nd digit has been selected, we cannot select it for the 3rd digit
So, we can accomplish stage 3 in 9 ways.

Stage 4: Select the 4th digit (the tens digit)
Once the 3rd digit has been selected, we cannot select it for the 4th digit
So, we can accomplish stage 4 in 9 ways.

Stage 5: Select the 5th digit (the units digit)
Once the 4th digit has been selected, we cannot select it for the 5th digit
So, we can accomplish stage 5 in 9 ways.

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus create a 5-digit number) in (9)(9)(9)(9)(9) ways ([spoiler]= 9^5 ways = C[/spoiler])

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by Pratiti » Sat May 18, 2013 6:38 am

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by jaspreetsra » Sat Dec 27, 2014 1:30 am
I just guessed C.
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by Brent@GMATPrepNow » Sat Dec 27, 2014 10:52 am
As I showed in my earlier solution (2 posts above), we can solve the question by applying the Fundamental Counting Principle (FCP).

NOTE: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

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by mendezsk » Mon Jul 06, 2015 3:47 pm
Hello All,

Please be of guidance. I am actually receiving D (9*8^4) as my answer.

Logic:

The question states how many 5 digit positives integers exist where no 2 consecutive integers are the same?

the 1st integer: 1-9
2nd integer: 8 possibilities (lets say the 1st integer was 2... the 2nd integer cannot be 2, therefore it can be 1,3,4,5,6,7,8, or 9 = 8 possibilities. Why are we including 0-9 when 0 is not a positive integer)?
3rd integer: 8 poss
4th integer: 8 poss
5th integer: 8 poss

9*8^4
That's what I am getting. So please help

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by Brent@GMATPrepNow » Mon Jul 06, 2015 3:52 pm
mendezsk wrote:Hello All,

Please be of guidance. I am actually receiving D (9*8^4) as my answer.

Logic:

The question states how many 5 digit positives integers exist where no 2 consecutive integers are the same?

the 1st integer: 1-9
2nd integer: 8 possibilities (lets say the 1st integer was 2... the 2nd integer cannot be 2, therefore it can be 1,3,4,5,6,7,8, or 9 = 8 possibilities. Why are we including 0-9 when 0 is not a positive integer)?
3rd integer: 8 poss
4th integer: 8 poss
5th integer: 8 poss

9*8^4
That's what I am getting. So please help
You're correct when you say that 0 is not a positive INTEGER.
BUT, it's okay to include 0 as one of the DIGITS in a positive INTEGER.

For example, 50532 is a positive integer that contains the non-positive DIGIT 0.

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by nikhilgmat31 » Sun Aug 16, 2015 11:19 pm
This question is different from other counting question.

9 * 9 * 9 * 9 * 9 = 9^5