2025 AMC 8 Problem 16

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To help you master 2025 AMC 8 Problem 16, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

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2025 AMC8 Problem 16

Solution

Solution 1

Pairing Method

Group the 20 available numbers into 10 pairs that differ by 10. Since we choose 10 numbers total and cannot pick both from any pair, we must choose exactly one from each pair. The sum is the sum of all base values plus the ‘bonuses’ from the larger numbers.

When distributing $N$ selections among $N$ restricted groups with a maximum of 1 selection per group, every group must be selected exactly once.

Step 1: Group the numbers into restricted pairs

The numbers from 1 to 20 can be organized into pairs where the two numbers differ by exactly 10. Any valid set of choices cannot contain both numbers from any of these pairs.

(1, 11), (2, 12), (3, 13), \dots, (10, 20)

Step 2: Determine the number of selections from each pair

We must select 10 numbers in total. The rules state no two chosen numbers can differ by exactly 10, meaning we cannot select both numbers from any pair. Because there are exactly 10 pairs available and we need to choose 10 numbers, we are forced to choose exactly one number from every single pair.

Step 3: Express the chosen numbers by their base values

Every pair consists of a ‘base value’ from 1 to 10, and a ‘larger value’. If we pick the base value, it contributes itself to the total sum. If we pick the larger value, it contributes the base value plus an extra 10.

Larger Value = Base Value + 10

Step 4: Set up the total sum calculation

Since we pick exactly one number from each of the 10 pairs, our total sum will include every base value from 1 to 10 exactly once. In addition, the problem states we chose exactly 5 numbers from the larger range (11 to 20). Each of these 5 choices adds an extra 10 to its base value.

Total Sum = (1 + 2 + \dots + 10) + 5 \times 10

Step 5: Calculate the final sum

Evaluate the arithmetic to find the final result. The sum of the first 10 positive integers is 55. Adding 50 yields the total sum.

Total Sum = 55 + 50 = 105

Correct answer: C. $105$
The sum is 105, which corresponds to answer choice C. We can verify this with a test case: let the 5 numbers from 1 to 10 be 1, 2, 3, 4, 5. To avoid a difference of 10, we cannot choose 11, 12, 13, 14, 15. Thus, we must choose 16, 17, 18, 19, 20. The sum is (1+2+3+4+5) + (16+17+18+19+20) = 15 + 90 = 105, confirming our result.

Solution 2

Algebraic Shift Method

Subtract 10 from the chosen numbers in the 11-20 range to transform them into numbers in the 1-10 range. This shows they perfectly complement the other chosen numbers to form a complete set from 1 to 10.

Transforming a set of variables by a constant offset can reveal hidden non-overlapping properties.

Step 1: Define sets of chosen numbers

Represent the 5 chosen numbers from the smaller range as one set, and the 5 chosen numbers from the larger range as a second set.

A = {a_{1}, a_{2}, a_{3}, a_{4}, a_{5}}

B = {b_{1}, b_{2}, b_{3}, b_{4}, b_{5}}

Step 2: Shift the larger numbers down

Subtract 10 from each number in the larger group to create a new shifted group. Because all original numbers were between 11 and 20, the new shifted numbers will all be between 1 and 10.

B^{'} = {b_{1} - 10, b_{2} - 10, b_{3} - 10, b_{4} - 10, b_{5} - 10}

Step 3: Prove the two groups perfectly combine

The problem states no two chosen numbers differ by exactly 10, which means no number in the original larger group can be exactly 10 more than any number in the smaller group. Consequently, no number in the shifted group can equal any number in the smaller group. Since the two groups together contain 10 distinct integers between 1 and 10, they must collectively contain every integer from 1 to 10 exactly once.

A \cup B^{'} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Step 4: Rewrite the total sum using the shift

The overall sum is the sum of the original smaller group plus the sum of the original larger group. We can rewrite the sum of the larger group in terms of the shifted group by adding back the 10 that was subtracted from each of its 5 elements.

Sum (A) + Sum (B) = Sum (A) + Sum (B^{'}) + (5 \times 10)

Step 5: Substitute the known total

Since the union of the smaller group and the shifted group contains all integers from 1 to 10 exactly once, their combined sum is simply the sum of 1 through 10. We substitute this known value and add 50 to find the total.

Total Sum = (1 + 2 + \dots + 10) + 50 = 55 + 50 = 105

Correct answer: C. $105$
The total sum is exactly 105 (Choice C). The algebraic transformation mathematically guarantees that the sum remains constant, regardless of which specific numbers are selected.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Sum of the First $n$ Integers:

i = 1 \sum n i = \frac{n ( n + 1 )}{2}

$n$ is the number of terms. Used to efficiently calculate $1 + 2 + \dots + 10 = \frac{10 \times 11}{2} = 55$ .

Pigeonhole Principle (Logical Variant):

If you have $N$ items to distribute into $N$ categories, and no category is allowed to contain more than 1 item, then every category must contain exactly 1 item.

Requires the number of items and categories to be equal, and a strict maximum of 1 item per category. Since 10 numbers are chosen (items) from 10 pairs (categories) without selecting two from any pair, exactly one number must be chosen from each pair.

Key Takeaway: Remember This Strategy for Next Time

Conditions restricting pairs of numbers, such as 'no two numbers differ by $X$ ' or 'no two sum to $Y$ '.

Try: Explicitly write out all the restricted pairs to see how selections in one group dictate forced choices in another.

Why: Pairing clarifies exactly which selections mutually exclude each other, transforming a complex condition into a simple rule: 'pick at most one from each pair'.

A problem involves choosing from two distinct sets that have a fixed offset (like 1 to 10 and 11 to 20).

Try: Shift one set algebraically (e.g., subtract the offset) so they can be compared directly within the same numerical range.

Why: Shifting reveals hidden relationships, showing how the subsets perfectly complement each other to cover all possible values without overlap.

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2025 AMC 8 Problem 16

In This Page:

2025 AMC8 Problem 16