2025 AMC 10A Problems/Problem 5

Consider the sequence of positive integers $1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2 \dots$

What is the $2025$ th term in the sequence?

$\textbf{(A) } 5 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 16 \qquad\textbf{(D) } 44 \qquad\textbf{(E) } 45$

Video Solution

https://youtu.be/l1RY_C20Q2M

Solution 1

One possible way the sequence could've been constructed was by putting "mountains" going up from $1$ , to $n+1,$ then going back down to $2.$ For example, the first few "mountains" look like this:

$12|1232|123432|12345432|...$

So, the $n^{th}$ mountain has length $2n$ and has highest number $n+1.$ We want to add mountains until we get a total length as close as possible, but not exceeding, $2025.$ Let the last mountain we sum be mountain $a.$ Hence, $2+4+6+...+2a=2(1+2+3+...+a)=a(a+1)\le2025$ $\implies a^2<2025\implies a<45,$ so our max $a$ is $44.$ In this $44^{th}$ mountain, the max number is $45,$ so the $45^{th}$ mountain has max number $46.$ Next, $44(44+1)=1980,$ so we're looking for the $45^{th}$ number in the $45^{th}$ mountain, which is $\boxed{\text{(E) }45}.$

~Tacos_are_yummy_1

Solution 2 (nice equation)

Group the numbers by their hill pattern: $(12)(1232)(123432)(12345432)...$

The maximums of each hill occur at terms $n = 2, 5, 10, 17...$ These terms correspond to maximums of $2, 3, 4,...$ Let $a$ be the maximum at term $N$ . Since the sum of the first $x$ odd numbers is $x^2$ we have $1 + (a-1)^2 = N.$ So for example, if a = $4,$ then $N = 10,$ telling us that the peak of the hill with maximum $4$ occurs at the $10$ th term.

Now, we know $2025 = 45^2,$ so let $a = 46.$ Then $N = 2026,$ so the $2026$ th term is $46.$ Then the $2025$ th term must be $\boxed{\text{(E) }45}.$

~grogg007

Solution 3

Note that the first $1$ is the first number from the left, the second $1$ is the third number from the left, and the third $1$ is the seventh number from the left. Since the second differences between the indices of the $n^{th}$ $1$ ’s are equal, there is a quadratic function for the position of the $n^{th}$ $1$ . As the second differences are $2$ , the leading coefficient of this function is $1$ . Let the function be $f(x)=x^2+ax+b$ . Then, we have $1+a+b=1$ , and $4+2a+b=3$ . Solving, we get $a=-1$ and $b=1$ . Therefore, the $n$ th $1$ from the left is the $(n^2-n+1)$ th number from the left. Plugging in $n=45$ gives $f(n)=1981$ , so the $1981$ st number is a $1$ .

Also, the sequence shows that the first instance of a positive integer $k$ is always to the left of the $k^{th}$ $1$ . This means that following the $1981^{st}$ number, the numbers go to or above $45$ . Adding $2025-1981$ to $1$ means that the $2025^{th}$ term of the sequence is $\boxed{\textbf{(E) }45}.$

~joiceeliu

~minor $\LaTeX$ edits by i_am_not_suk_at_math (saharshdevaraju 11:09, 7 November 2025 (EST)saharshdevaraju)

Solution 4 (Quick and Fast!🚀)

Note that after every round, the number of terms is the highest term squared. For example, after the threes, there are $3^2 = 9$ terms. We also notice that $45^2 = 2025$ , so the $2025$ th term is $45$ , which is answer choice $\boxed{\textbf{(E)}45}.$

~iiiiiizh

~minor $\LaTeX$ edits by i_am_not_suk_at_math (saharshdevaraju 09:35, 7 November 2025 (EST)saharshdevaraju)

Solution 5 (Super Simple!!)

Upon inspecting the problem, we see that the $1^{st}$ term is 1, and so are the $3^{rd}$ , $7^{th}$ , $13^{th}$ , $18^{th}$ , $31^{st}$ , and so on. We see that the pattern comes from the differences: $+2$ , $+4$ , $+6$ , $+8$ , $+10$ , and so on. From this we see that a $1$ appears at each position that is a number of the form $n^{2}-n+1$ . We know that $2025 = 45 \cdot 45$ . Plugging $45$ as $n$ gives $1981$ , meaning that the $2025^{th}$ term is $2025-1981+1=45$ . Therefore, the solution is $\boxed{\textbf{(E)}45}.$

~vgarg

~major $\LaTeX$ edits by i_am_not_suk_at_math (saharshdevaraju 15:27, 7 November 2025 (EST)saharshdevaraju)

Solution 6 - ⚡ Fastest and Easiest! ⚡

$\large \textbf{Takes less than 30 seconds!}$

Immediately by looking at the sequence, we observe that the digit in the place of each perfect square is the square root of that number. For example, the first term in the sequence is $1$ (because $\sqrt{1}=1$), the $4^{th}$ term is $2$ (because $\sqrt{4}=2$), the $9^{th}$ term is $3$ (because $\sqrt{9}=3$), the $16^{th}$ term is $4$ (because $\sqrt{16}=4$), and so on. And since $2025$ is a perfect square, it makes it easy for us because $\sqrt{2025}=45$, which makes the answer choice $\boxed{\textbf{(E) }45}.$

To make it easier to understand my way of solving it, here is a table.

Let $n =$ the position of the said number in the sequence

Let $\sqrt{n} =$ the number located in that position itself

\begin{array}{c|c|c} n & \sqrt{n} \\[6pt] \hline 1 & 1 \\[6pt] 4 & 2 \\[6pt] 9 & 3 \\[6pt] 16 & 4 \\[6pt] 25 & 5 \\[6pt] 36 & 6 \\[6pt] \dots & \dots \\[6pt] 2025 & \boxed{\textbf{(E) }45} \end{array}

$\textbf{Note}$: I am just saying that the square root of the position (if the position is a perfect square) is the number that appears in the said position in the above sequence. I am not saying that the root itself is the first time it appears in the said sequence. To elaborate, the number $6$ appears on the $26^{th}$ number, as $\textit{well}$ as the $36^{th}$ number as expected (because $\sqrt{36} = 6$).

~i_am_not_suk_at_math (saharshdevaraju 21:14, 6 November 2025 (EST)saharshdevaraju)

Solution 7 - Pattern Recognition

After looking at the problem, it is evident that a pattern exists for the first occurence of a number in the sequence. Knowing this, we can write the places that each number appears at: \begin{align} &1: 1 \\ &2: 2 \\ &3: 5 \\ &4: 10 \\ &5: 17 \end{align} If we write down the difference between the each numbers position in sequence, we get $1, 3, 5, \text{and } 7$ . We notice how these numbers are similar to how the square numbers increase (square numbers: $1, 4, 9, 16…$ ). We also notice how the position of the first instance of each number $n$ is just $(n-1)^2+1$ . Since we're trying to find the $2025^{th}$ number and $2025 = 45^2$ , we can plug in $45$ for $n-1$ . As a result, we get that the first instance of the number $46$ appears at $(45)^2+1$ or $2026$ . Knowing that the number before the first instance of a number $n$ is $n-1$ , we get the answer choice $\boxed{\textbf{(E) }45}.$

~Kasoisanti

~minor edits by i_am_not_suk_at_math (saharshdevaraju 09:34, 7 November 2025 (EST)saharshdevaraju)

~minor $\LaTeX$ edits by i_am_not_suk_at_math (saharshdevaraju 15:22, 7 November 2025 (EST)saharshdevaraju)

Solution 8 - Simple solution for dummies

We can list out all of the 2025 terms. If each term takes about 1 second, then it will take about 34 minutes in total, which is good enough if you are aiming for a 42 on the AMC 10. After listing all these terms, we can get the answer choice $\boxed{\textbf{(E) }45}.$

~dingpanda

~this is actually really funny -i_am_not_suk_at_math

Solution 9: Patterns

We list the positions of the $\text{1s}$ in the sequence and get that they appear at $1,3,7,13,...$ We see that the $n\text{th}$ $1$ in the sequence, its location is at the $n(n+1)+1\text{th}$ position. So, we can find that the $1$ closest to $2025$ happens where $n=44,$ due to our knowledge that $2025=45^2.$ Since $44(45)+1=1981,$ we can start counting from $1981,$ getting that the 2025th term in the sequence is $\boxed{45}.$

~ gogogo2022

Solution 10: Sum of Series

We have the following series given by the question: $1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1 \dots$

Noting the differences between two consecutive numbers $a_{i+1} - a_i$ gives us the following:

$[1, -1], [1, 1, -1, -1], [1, 1, 1, -1, -1, -1], [1, 1, 1, 1, -1, -1, -1, -1], [1, 1, 1, 1, -1, -1, -1, -1]\dots$

Notice that the problem now becomes finding the sum of the first 2025 numbers, instead of the 2025th number. We also notice that every $1$ gets canceled eventually, and from inspection, the last $1$ in a consecutive string of $n$ $1$ s is always corresponding to the number at position $n^2.$ For example, the 9th number is the last $1$ in the $\sqrt{9} = 3$ consecutive numbers, and since the numbers before these consecutive $1$ s cancel out, the sum ( $3$ ) is the 9th number in the original series. So our answer must be $\sqrt{2025} = \boxed{\textbf{(E) }45}.$

~ Cheetahboy93

Chinese Video Solution

https://www.bilibili.com/video/BV1gV2uBbEJe/

~metrixgo

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

Video Solution (Fast and easy)

https://youtu.be/ZQhAdIs2FIg?si=mEgnYvdQ2sMrCx1P ~ Pi Academy

Video Solution by Daily Dose of Math

https://youtu.be/LN5ofIcs1kY

~Thesmartgreekmathdude

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2025 AMC 10A Problems/Problem 5

Contents

Video Solution

Solution 1

Solution 2 (nice equation)

Solution 3

Solution 4 (Quick and Fast!🚀)

Solution 5 (Super Simple!!)

Solution 6 - ⚡ Fastest and Easiest! ⚡

Solution 7 - Pattern Recognition

Solution 8 - Simple solution for dummies

Solution 9: Patterns

Solution 10: Sum of Series

Chinese Video Solution

Video Solution by SpreadTheMathLove

Video Solution

Video Solution (Fast and easy)

Video Solution by Daily Dose of Math

See Also