Dividing 12 by 3 gives 4 as each equal part. This is a basic division problem that checks understanding of equal partitioning.

Dividing 20 by 5 gives 4 as the value of each part. This question reinforces the idea of equal partitioning with whole numbers.

Dividing a whole into 4 equal parts means each friend gets 1/4 of the cake. This problem introduces partitioning in the context of fractions.

Since 5 + 7 equals 12, the missing part x must be 15 - 12, which is 3. The question assesses basic subtraction in a partitioning context.

When a whole is divided into 8 equal parts, each part represents 1/8 of the whole. This reinforces the concept of equal fractional parts.

Let the numbers be x, x+2, and x+4. Setting up the equation x + (x+2) + (x+4) = 30 results in 3x + 6 = 30, so x = 8. This checks basic algebra with consecutive even numbers.

If one part is twice the other, set the parts as x and 2x. Their sum being 12 leads to 3x = 12, so x = 4 and the parts are 4 and 8. This reinforces setting up equations based on ratio relationships.

Dividing the total area 36 by 3 gives an area of 12 for each part. The problem uses geometric partitioning to enhance spatial reasoning in mathematics.

The ratio totals 1+2+3 = 6, and the middle part has a value of 2, so it represents 2/6 which simplifies to 1/3 of the total. This question combines ratio reasoning with partitioning.

Dividing 1/2 into two equal parts gives 1/4 for each part. This exercise helps solidify the understanding of partitioning fractions.

The segments are 2, 3, 4, 5, and 6. Their sum is 20, which represents the total length of the bar. This question applies arithmetic series concepts in a partitioning context.

Let the smaller group be x; then the larger group is 2x. With x + 2x = 21, we have x = 7, so the groups are 7 and 14. This reinforces proportional reasoning in partition problems.

Let the second part be x. Then the parts are (x-5), x, and (x+5). Their sum is 3x, which equals 45, so x is 15. This problem uses simple algebra in a partitioning scenario.

Let x be the number of groups with 8 students, then (8 - x) groups have 9 students. Setting up the equation 8x + 9(8 - x) = 72 yields x = 0, meaning every group has 9 students. This challenges students to set up and solve a simple equation.

The distinct partitions of 6 into two parts (ignoring order) are 1+5, 2+4, and 3+3, totaling 3. This problem reinforces basic number partition concepts with small integers.

The distinct partitions of 7 into three parts, without considering order, are {1,1,5}, {1,2,4}, {1,3,3}, and {2,2,3}. This enumeration results in 4 unique partitions.

Using the geometric series sum 2(1 + r + r²) = 30 leads to the equation r² + r - 14 = 0. Solving this gives a positive value of approximately 3.28, which rounds to about 3.3.

Among the options, only 2 + 7 + 11 equals 20 and all numbers are prime and distinct. This problem challenges students to combine number theory with partitioning concepts.

Enumerating partitions of 10 into distinct parts yields 1 partition with one number, 4 with two numbers, 4 with three numbers, and 1 with four numbers, totaling 10 partitions. This exercise deepens understanding of partition functions.

Using the sum formula for an arithmetic series, the total length is given by 5n(n+1)/2 = 100, which simplifies to n(n+1) = 40. Since no positive integer n satisfies this equation, there is no solution for n.