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Kobotan [32]
3 years ago
9

A small combination lock on a suitcase has 55 ​wheels, each labeled with the 10 digits 0 to 9. How many 55 digit combinations ar

e possible if successive digits must be​ different?
Mathematics
1 answer:
den301095 [7]3 years ago
7 0
Suppose a_n is the number of possible combinations for a suitcase with a lock consisting of n wheels. If you added one more wheel onto the lock, there would only be 9 allowed possible digits you can use for the new wheel. This means the number of possible combinations for n+1 wheels, or a_{n+1} is given recursively by the formula

a_{n+1}=9a_n

starting with a_1=10 (because you can start the combination with any one of the ten available digits 0 through 9).

For example, if the combination for a 3-wheel lock is 282, then a 4-wheel lock can be any one of 2820, 2821, 2823, ..., 2829 (nine possibilities depending on the second-to-last digit).

By substitution, you have

a_{n+1}=9a_n=9^2a_{n-1}=9^3a_{n-2}=\cdots=9^na_1=10\times9^n

This means a lock with 55 wheels will have

a_{55}=10\times9^{54}

possible combinations (a number with 53 digits).
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First simplify the rational expression by dividing. The degree in the numerator has to be at least 1 less than the degree in the denominator before you can decompose into partial fractions.

(3<em>x</em>³ - 5<em>x</em>² - 3<em>x</em> - 40) / ((<em>x</em>² + 4) (<em>x</em> - 3)) = 3 + (4<em>x</em>² - 15<em>x</em> - 4) / ((<em>x</em>² + 4) (<em>x</em> - 3))

Now decompose the remainder term into partial fractions:

(4<em>x</em>² - 15<em>x</em> - 4) / ((<em>x</em>² + 4) (<em>x</em> - 3)) = (<em>ax</em> + <em>b</em>) / (<em>x</em>² + 4) + <em>c</em> / (<em>x</em> - 3)

Multiply both sides by the denominator on the left:

4<em>x</em>² - 15<em>x</em> - 4 = (<em>ax</em> + <em>b</em>) (<em>x</em> - 3) + <em>c</em> (<em>x</em>² + 4)

Expand the right side:

4<em>x</em>² - 15<em>x</em> - 4 = <em>ax</em>² + (<em>b</em> - 3<em>a</em>) <em>x</em> - 3<em>b</em> + <em>cx</em>² + 4<em>c</em>

4<em>x</em>² - 15<em>x</em> - 4 = (<em>a</em> + <em>c</em>) <em>x</em>² + (<em>b</em> - 3<em>a</em>) <em>x</em> - 3<em>b</em> + 4<em>c</em>

Then

<em>a</em> + <em>c</em> = 4

<em>b</em> - 3<em>a</em> = -15

-3<em>b</em> + 4<em>c</em> = -4

Solve this system to get

<em>a</em> = 5, <em>b</em> = 0, <em>c</em> = -1

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(4<em>x</em>² - 15<em>x</em> - 4) / ((<em>x</em>² + 4) (<em>x</em> - 3)) = 5<em>x</em> / (<em>x</em>² + 4) - 1 / (<em>x</em> - 3)

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