There are four even integers in the top five rows of Pascal's Triangle. How many even integers are in the top 10 rows of the triangle?
We can list the first 10 rows of Pascal's triangle, and mark the even numbers.

[asy]
usepackage("amsmath");
unitsize(0.5 cm);

int i, j, n;

for (int i = 0; i <= 9; ++i) {
for (int j = 0; j <= 9; ++j) {
  if (i + j <= 9) {
    n = choose(i + j,i);
    if (n % 2 == 0) {label("$\boxed{" + string(n) + "}$", i*(-1,-1) + j*(1,-1));}
    if (n % 2 == 1) {label("$" + string(n) + "$", i*(-1,-1) + j*(1,-1));}
  }
}}
[/asy]

Thus, the number of even numbers is $1 + 3 + 2 + 3 + 7 + 6 = \boxed{22}.$
22
Level 5
Counting & Probability
Problem: There are four even integers in the top five rows of Pascal's Triangle. How many even integers are in the top 10 rows of the triangle?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the number of even integers in the top 10 rows of Pascal's Triangle, we first need to understand the structure of Pascal's Triangle. Each entry in the triangle is given by the binomial coefficient C(n, k), which represents the number of ways to choose k elements from a set of n elements. The first 10 rows of Pascal's Triangle correspond to n values from 0 to 9.

Next, we calculate the binomial coefficients for each row:

Row 0: C(0, 0) = 1  
Row 1: C(1, 0) = 1, C(1, 1) = 1  
Row 2: C(2, 0) = 1, C(2, 1) = 2, C(2, 2) = 1  
Row 3: C(3, 0) = 1, C(3, 1) = 3, C(3, 2) = 3, C(3, 3) = 1  
Row 4: C(4, 0) = 1, C(4, 1) = 4, C(4, 2) = 6, C(4, 3) = 4, C(4, 4) = 1  
Row 5: C(5, 0) = 1, C(5, 1) = 5, C(5, 2) = 10, C(5, 3) = 10, C(5, 4) = 5, C(5, 5) = 1  
Row 6: C(6, 0) = 1, C(6, 1) = 6, C(6, 2) = 15, C(6, 3) = 20, C(6, 4) = 15, C(6, 5) = 6, C(6, 6) = 1  
Row 7: C(7, 0) = 1, C(7, 1) = 7, C(7, 2) = 21, C(7, 3) = 35, C(7, 4) = 35, C(7, 5) = 21, C(7, 6) = 7, C(7, 7) = 1  
Row 8: C(8, 0) = 1, C(8, 1) = 8, C(8, 2) = 28, C(8, 3) = 56, C(8, 4) = 70, C(8, 5) = 56, C(8, 6) = 28, C(8, 7) = 8, C(8, 8) = 1  
Row 9: C(9, 0) = 1, C(9, 1) = 9, C(9, 2) = 36, C(9, 3) = 84, C(9, 4) = 126, C(9, 5) = 126, C(9, 6) = 84, C(9, 7) = 36, C(9, 8) = 9, C(9, 9) = 1  

Now, we identify the even integers in each row:

Row 0: 0 even integers  
Row 1: 0 even integers  
Row 2: 1 even integer (2)  
Row 3: 0 even integers  
Row 4: 3 even integers (4, 6, 4)  
Row 5: 3 even integers (10, 10)  
Row 6: 3 even integers (6, 20, 6)  
Row 7: 2 even integers (21 is odd)  
Row 8: 5 even integers (8, 28, 56, 70, 8)  
Row 9: 6 even integers (36, 84, 126, 126, 84)  

Now we sum the even integers from each row:

Row 0: 0  
Row 1: 0  
Row 2: 1  
Row 3: 0  
Row 4: 3  
Row 5: 3  
Row 6: 3  
Row 7: 2  
Row 8: 5  
Row 9: 6  

Total even integers = 0 + 0 + 1 + 0 + 3 + 3 + 3 + 2 + 5 + 6 = 22.

Thus, the final answer is 22.<|END_OF_LATENT|> The final answer is $22$.