The first six rows of Pascal's triangle are shown below, beginning with row zero. Except for the $1$ at each end, row $4$ consists of only even numbers, as does row $2.$ How many of the first $20$ rows have this property? (Don't include row $0$ or row $1$). \begin{tabular}{ccccccccccc} &&&&&1&&&&&\\ &&&&1&&1&&&&\\ &&&1&&2&&1&&&\\ &&1&&3&&3&&1&&\\ &1&&4&&6&&4&&1&\\ 1&&5&&10&&10&&5&&1\\ \end{tabular}
+505 
Notice that only 2^n rows (where n is a positive integer) can obtain this property, where the row only contains even numbers except for the beginning and the end. (For example, only row 2^1 = 2, row 2^2=row 4, row 2^3=row 8, and so on). There are 4 numbers that can be expressed as 2^n (where n is a positive integer) under 20 (which are 2, 4, 8, and 16), so the answer is \(\boxed{4}\).
My intuition for the fact that only 2^n rows (where n is a positive integer) can obtain this property uses the fact that the numbers in the pascal triangle can be expressed as \(\frac{n!}{p!(n-p)!}\), where n is the nth row and p is the pth number in that row. All odd numbers obviously don't work, because if p = 1, then the result is just going to be equal to n, and since n is odd, the pascal triangle contains odd numbers. Even numbers that cannot be expressed as 2^n (where n is an integer) also cannot have only even numbers, because it must also have odd factors, and if it has an odd factor, a certain value of p will cancel out all the even factors and leave the number odd, so only 2^n would work.
Also a very similar question was answered a long time ago: https://web2.0calc.com/questions/another-pascal-s-triangle-question
+505
Notice that only 2^n rows (where n is a positive integer) can obtain this property, where the row only contains even numbers except for the beginning and the end. (For example, only row 2^1 = 2, row 2^2=row 4, row 2^3=row 8, and so on). There are 4 numbers that can be expressed as 2^n (where n is a positive integer) under 20 (which are 2, 4, 8, and 16), so the answer is \(\boxed{4}\).
My intuition for the fact that only 2^n rows (where n is a positive integer) can obtain this property uses the fact that the numbers in the pascal triangle can be expressed as \(\frac{n!}{p!(n-p)!}\), where n is the nth row and p is the pth number in that row. All odd numbers obviously don't work, because if p = 1, then the result is just going to be equal to n, and since n is odd, the pascal triangle contains odd numbers. Even numbers that cannot be expressed as 2^n (where n is an integer) also cannot have only even numbers, because it must also have odd factors, and if it has an odd factor, a certain value of p will cancel out all the even factors and leave the number odd, so only 2^n would work.
Also a very similar question was answered a long time ago: https://web2.0calc.com/questions/another-pascal-s-triangle-question
