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Difficult Math Problem #112 - Number Theory

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Difficult Math Problem #112 - Number Theory

by 800guy » Thu Mar 29, 2007 10:37 pm
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


oa coming when a few people respond with explanations. from difficult math doc.
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Re: Difficult Math Problem #112 - Number Theory

by gabriel » Fri Mar 30, 2007 7:08 am
800guy 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


oa coming when a few people respond with explanations. from difficult math doc.

excluding 0 the first digit can be written in 9 ways .... excluding the first digit the second digit can be written in 9 ways ... so on.... the answer is 9^5 .. C
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Re: Difficult Math Problem #112 - Number Theory

by Prasanna » Sat Mar 31, 2007 4:44 am
gabriel wrote:
800guy 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


oa coming when a few people respond with explanations. from difficult math doc.

excluding 0 the first digit can be written in 9 ways .... excluding the first digit the second digit can be written in 9 ways ... so on.... the answer is 9^5 .. C
Hi Gabriel

I have a doubt. The question says that the no two consecutive digits are the same. If the second digit can be written in 9 ways, is there not a chance for repeating the first number. (Sorry if my question was stupid)

Prasanna
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Re: Difficult Math Problem #112 - Number Theory

by jayhawk2001 » Sat Mar 31, 2007 6:42 am
Prasanna wrote:
I have a doubt. The question says that the no two consecutive digits are the same. If the second digit can be written in 9 ways, is there not a chance for repeating the first number. (Sorry if my question was stupid)

Prasanna
For the first digit, we are excluding 0. For the second digit, we are
including 0 and excluding the first digit (as it reappears). So, the number
of ways remain at 9
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by Prasanna » Sat Mar 31, 2007 7:31 am
Got it. I did not consider zero and hence went wrong. Thanks for the explanation.
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oa

by 800guy » Mon Apr 02, 2007 12:03 pm
oa:

C is correct.
The first place has 9 possibilities, since 0 is not to be counted. All others have 9 each, since you cannot have the digit, which is same as the preceding one.
Hence 9^5
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Re: oa

by MJay » Tue Apr 03, 2007 4:18 am
800guy wrote:oa:

C is correct.
The first place has 9 possibilities, since 0 is not to be counted. All others have 9 each, since you cannot have the digit, which is same as the preceding one.
Hence 9^5
Hi All,

I have a small doubt in the solution. Should we consider consecutive numbers in the ascending order alone?

To explain my point further ... Say the 1st place was 7, the next cannot take 6 and 8 because 7-8 are consecutive digits and 7-6 are also consecutive but in the reverse order. In this case the solution would be "the first place can be filled in 9 ways excluding zero, after which the rest of the places can be filled in only 8 ways"

Should this option be considered? I have interpreted "consecutive" in a literal sense to refer to the immediately preceeding and following digits.

I am a newbie here ... so do excuse me if my query sounds silly.

Regards,
MJay
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by Neo2000 » Tue Apr 03, 2007 8:11 am
When you say consecutive it always means the next number. So if you put 7, 8 was your next number
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