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2025 AMC 10A Problems/Problem 5: Difference between revisions

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-Kasoisanti
-Kasoisanti
==Chinese Video Solution==
https://www.bilibili.com/video/BV1gV2uBbEJe/
~metrixgo


==Video Solution==
==Video Solution==

Revision as of 01:45, 7 November 2025

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

  • This problem and problem 11 should've switched places tbh ;-; ~tacos
  • FRR

Video Solution (Fast and easy)

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

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 $\boxed{\text{(E) }45}.$

~joiceeliu

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 $\boxed{\text{(E)}45}.$

~iiiiiizh

Solution 5 (Super Simple!!)

Upon inspecting the problem, we see that the 1st term is 1, and so are the 3rd, 7th, 13th, 18th, 31st, 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*45. Plugging 45 as n gives 1981, meaning that the 2025th term is 2025-1981+1=45. Therefore, the solution is $\boxed{\text{(E)}45}.$

~vgarg


Solution 6 - ⚡ Very Fast! - Basic Math (& Pattern Recognition)

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 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.

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

~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 $\boxed{\text{(E) }45}.$

-Kasoisanti

Chinese Video Solution

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

~metrixgo

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

Video Solution by Daily Dose of Math

https://youtu.be/LN5ofIcs1kY

~Thesmartgreekmathdude

See Also

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

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