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Difference between revisions of "2008 iTest Problems/Problem 78"

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This gives an answer of <math>165-12=153</math>.
 
This gives an answer of <math>165-12=153</math>.
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==See Also==
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{{2008 iTest box|num-b=77|num-a=79}}
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 04:14, 28 January 2019

Problem

Feeling excited over her successful explorations into Pascal's Triangle, Wendy formulates a second problem to use during a future Jupiter Falls High School Math Meet:

How many of the first 2010 rows of Pascal's Triangle (Rows 0 through 2009) have exactly 256 odd entries?

What is the solution to Wendy's second problem?

Solution

Building off of Method 2 in problem 77, we see that each that the base 2 representation of the number reveals to us how many times we end up applying the recursion relation $O_k=2*O_{k-2^{n-1}}$.

For example, $2008=1024+512+256+128+64+16+8=(11111011000)_2$ has seven nonzero terms in its base 2 expansion so that we end up applying the recursion relation seven times giving a final term of $O_{2008}=2^7\cdot O_{0}=2^7=128$. So we need to count the number of integers between $\{0,2009\}$ which have exactly 8 nonzero terms in their base 2 expansion.

$2009=(11111011001)_2$.

$\binom{11}{8}$ counts the numbers up to $2047=2^{11}-1$ which have exactly 8 nonzero terms in their base 2 expansion. Of these $\binom{11}{8}=165$ candidates, the following 12 must be excluded since they exceed $2009$:

$2040=(11111111000)_2,2036(11111110100)_2,$

$2034=(11111110010)_2,2033=(11111110001)_2,$

$2028=(11111101100)_2,2026=(11111101010)_2,$

$2025=(11111101001)_2,2022=(11111100110)_2,$

$2021=(11111100101)_2,2019=(11111100011)_2,$

$2012=(11111011100)_2,2010=(1111011010)_2$

This gives an answer of $165-12=153$.


See Also

2008 iTest (Problems)
Preceded by:
Problem 77 		Followed by:
Problem 79
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