sat ii math practice problem

[lyzhou1990]

Hello,
I came across one of these problems in yet another of my forays into the realm of practice tests for subject tests. Here it is:

35. How many 5 digit integers do not contain successive digits that are the same?
A) 10^5
B) 9^5
C) 9 x 10^4
D) (9^2) x 8 x 7 x 6
E) 9 x 8 x 7 x 6 x 5

My problem is that I have no idea how to figure this out without using a "brute force" method (i.e. counting them out...one by one by one, FOREVER).

The book says that the answer is B, and gave the following cryptic explanation:

~~~~~~
9 9 9 9 9
| |
| |---> cannot be the same as the immediate left hand neighbor
|
|----> 0 cannot be the leftmost digit
~~~~~~

I have no idea what this means, but would like to know, so that the next time I come across a similar problem I won't have to resort to the tiresome brute-force method again.

Thanks.

 

[TchrWill]

Hello,
I came across one of these problems in yet another of my forays into the realm of practice tests for subject tests. Here it is:

35. How many 5 digit integers do not contain successive digits that are the same?
A) 10^5
B) 9^5
C) 9 x 10^4
D) (9^2) x 8 x 7 x 6
E) 9 x 8 x 7 x 6 x 5

My problem is that I have no idea how to figure this out without using a "brute force" method (i.e. counting them out...one by one by one, FOREVER).

The book says that the answer is B, and gave the following cryptic explanation:

~~~~~~
9 9 9 9 9
| |
| |---> cannot be the same as the immediate left hand neighbor
|
|----> 0 cannot be the leftmost digit
~~~~~~

I have no idea what this means, but would like to know, so that the next time I come across a similar problem I won't have to resort to the tiresome brute-force method again.

The way I interpret "How many 5 digit integers do not contain successive digits that are the same?" is how many of the 5 digit numbers from 10,000 to 99,999 do not contain 5 successive digits that are all ther same.

The only number containg 5 successive digits the same are 11,111, 22,222, 33,333, 44,444........99,999 or nine in all. Since there are 89,999 5 digit numbers, the answer to my interpretaion of your problem statement is 89,999 - 9 = 89,990.

If your problem statement really means something else, please let me know.

 

[galactus]

I believe what they mean is no 2 identical digits side-by-side.

For instance, 12345 or 12132 would be acceptable. The similar digits are

not side-by-side.

What you have is a classic counting problem. The first digit can not be 0, so you have 9 digits to choose from, 1-9

The next digit can be a 0, but not the same as the first digit. There are 9 to choose from.

The third digit can not be the same as the 2nd digit, but can be the same as the first and so on.

Therefore, you have 9*9*9*9*9=9^5=59049 possible combinations meeting the aforementioned criteria.

 

[Denis]

How many 5 digit integers do not contain successive digits that are the same?

---> cannot be the same as the immediate left hand neighbor
---> 0 cannot be the leftmost digit

To start, you need 5digit integers where there is NO 2 side by side digits
that are the same, like 12234.

So that clarifies "immediate left hand neighbor"; "22" means the 2nd "2"
has a "2" as immediate left hand neighbor...capish?

Also, 01234 is not acceptable, being really 4digit integer 1234; so that
clarifies the "0 cannot be the leftmost digit" rule.

The numbers will total 9^5, or 59049, like this:
1: 10101
2: 10102
...
59048: 98987
59049: 98989

And since there are 99999-10000+1 = 90000 total 5digit integers,
then there are 90000-59049 = 30,951 with at least 2 side by side
digits the same, like 12234, 12333 and the likes.

I'm not sure how to get these by "formula"; however, I think that
getting the 30,951 would be easier as 1st step.

It is quite easy to get the 59049 numbers using this simple looper:
(a,b,c,d,e are the 5 digits)

loop a from 1 to 9
loop b from 0 to 9 : skip if b=a
loop c from 0 to 9 : skip if c=b
loop d from 0 to 9 : skip if d=c
loop e from 0 to 9 : skip if e=d
count = count + 1