2025 AMC 8 Problem 13

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To help you master 2025 AMC 8 Problem 13, 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 13

Solution

Solution 1

Periodic Remainder Grouping Method

Find one complete cycle of remainders and then divide the $25$ even numbers into complete cycles plus a leftover group. This avoids dividing all $25$ numbers separately.

If two integers differ by a multiple of $7$, they have the same remainder when divided by $7$. Since seven consecutive even numbers span an increase of $14$, the remainder pattern repeats after every seven terms.

Step 1: Count the even numbers

The list is an arithmetic sequence beginning at $2$, ending at $50$, and increasing by $2$. Counting its terms shows how many remainder results the histogram must contain.

$$\frac{50-2}{2}+1=25$$

Step 2: Find one complete remainder cycle

Divide the first seven even numbers by $7$ and record only their remainders. Every possible remainder occurs exactly once in this group.

$$2,4,6,8,10,12,14$$

$$2,4,6,1,3,5,0(\mathrm{mod}7)$$

Step 3: Explain why the pattern repeats

The even number seven positions later is $14$ greater. Because $14$ is divisible by $7$, adding it does not change the remainder.

$$14=2\cdot 7$$

$$(n+14)\mathrm{mod}7=n\mathrm{mod}7$$

Step 4: Separate complete cycles from leftover terms

Three complete cycles use the first $21$ numbers, so each remainder occurs three times. The four remaining numbers follow the first four entries of the cycle, giving one additional occurrence to remainders $2,4,6,1$.

$$25=3\cdot 7+4$$

$$\text{extra remainders}=2,4,6,1$$

Step 5: Record the frequency of each remainder

Start each remainder with three occurrences and add one to each of the four extra remainders. Reading in the histogram’s order from remainder $0$ through remainder $6$ gives the required bar heights.

$$r=0:3$$

$$r=1:4$$

$$r=2:4$$

$$r=3:3$$

$$r=4:4$$

$$r=5:3$$

$$r=6:4$$

$$(3,4,4,3,4,3,4)$$

Correct answer: A.
The histogram with heights $(3,4,4,3,4,3,4)$ is answer choice $\boxed{\text{A}}$. As a check, the bar heights total $3+4+4+3+4+3+4=25$, matching the number of even integers in the original list.

Similar Problems in Real Exams

The following five AMC 8 problems are closely related to AMC 8 2025 Question 13. These real AMC 8 problems test the same key concept—Remainders & Modular Arithmetic. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2025 AMC 8 Problem 6

2019 AMC 8 Problem 14

2018 AMC 8 Problem 21

2016 AMC 8 Problem 5

2013 AMC 8 Problem 1

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Number of terms in an arithmetic sequence:

$F$ is the first term, $L$ is the last term, and $d$ is the constant positive difference; the endpoints must belong to the sequence. The even numbers from $2$ through $50$ form an arithmetic sequence with first term $2$, last term $50$, and common difference $2$.

• Remainders are unchanged by adding a multiple of the divisor:

$a$ and $b$ are integers, and $m$ is a positive divisor. Terms seven positions apart differ by $14$, which is a multiple of $7$, so their remainders are equal.

• Complete-cycle counting:

$N$ is the total number of terms, $k$ is the cycle length, $q$ is the number of complete cycles, and $0\le r<k$. The remainder cycle has length $7$, and $25=3\cdot 7+4$, so there are three complete cycles and four leftover remainders.

Key Takeaway: Remember This Strategy for Next Time

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