# Matrix and Binomial Coefficients

Considering the construction of a matrix as follows.

The $n$th row in the matrix is filled with the coeffcients of $x^r$ in the expansion of $(1+x)^n$ from the columns $2n$ to $3n$ inclusive and circle all the numbers that are divisible by $n$ in the same row

How would I find the number of columns for which all the elements in a column are circled in the first j columns given j = 547 ?

• "How would I find the number of columns for which all the elements in a column are circled in the first j columns given j = 547 ?" It feels that one of the "columns" must actually read "rows". Is the question correct? – Srivatsan Jul 27 '11 at 22:59

## 1 Answer

Hint: Have you tried looking at Pascal's triangle modulo some small primes? You could see a pattern that would apply. The Divisibility properties section of Wikipedia's Binomial Coefficient article has some useful information. This page has some neat images mod 2.

• Is the answer 547C2 where nCr = n!/(r!*(n-r)!) ? – user6349 Jan 28 '11 at 15:03
• Can you please provide some more insight? – user6349 Jan 28 '11 at 15:10
• It can't be that big. You only have 547 columns to consider. Start with 3's. In row 3, columns 6 through 9 have 1, 3, 3, and 1, and you circle the two 3's in 7 and 8. Row 4 has 1,4,6,4,1 in columns 8 through 12 and you circle the 4's in 9 and 11. So column 8 doesn't have all the numbers circled, but 7 does. If you make a spreadsheet going up to 20 or so you will probably see the pattern. – Ross Millikan Jan 28 '11 at 15:19
• Could I get an explanation for the downvote? – Ross Millikan Jan 28 '11 at 15:28
• Also, any row that is a multiple of 2 or 3 will have a 1 in it (like row 3 in my example), which you won't circle. So you only have to consider rows of the form 6k+1 and 6k+5. – Ross Millikan Jan 28 '11 at 20:51