AMC 10 · 2025 · #2

Easy mode Grade 5

Problem

Square each whole number from $1$ up to $2025$ . For every square, write down only its last digit (the ones digit). For example, $1^2=1$ gives $1$ , $2^2=4$ gives $4$ , $3^2=9$ gives $9$ , and $4^2=16$ gives $6$ . Add up all $2025$ of these last digits. What is the total?

Pick an answer.

(A)

9025

(B)

9070

(C)

9090

(D)

9115

(E)

9160

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: For each of the first $2025$ positive perfect squares $1^2,2^2,3^2,\dots,2025^2$, take only the ones digit. Add all $2025$ of those ones digits together and report the total.

Givens: Jerry lists the ones digit of $n^2$ for every $n$ from $1$ to $2025$; The listed digits begin $1,4,9,6,5,6,\dots$ (the ones digits of $1^2,2^2,3^2,4^2,5^2,6^2,\dots$); Answer choices: (A) $9025$, (B) $9070$, (C) $9090$, (D) $9115$, (E) $9160$

Unknowns: The sum of all $2025$ ones digits that Jerry wrote down

Understand

Restated: For each of the first $2025$ positive perfect squares $1^2,2^2,3^2,\dots,2025^2$, take only the ones digit. Add all $2025$ of those ones digits together and report the total.

Plan

Primary tool: #5 Look for a Pattern

Secondary: #7 Identify Subproblems

Adding $2025$ separate ones digits by hand is hopeless, so Tool #5 (Look for a Pattern) is the key move: the ones digit of $n^2$ is fixed by the ones digit of $n$, so the list of ones digits repeats in blocks of $10$. Once the repeating block is known, Tool #7 (Identify Subproblems) splits the count $2025$ into whole blocks plus a short leftover, turning one giant sum into a small multiplication plus a tiny add.

Execute — Answer: D

#5 Look for a Pattern 4.NBT.B.5 Step 1

Find the repeating block of ten

The ones digit of a product depends only on the ones digits of the factors, so the ones digit of $n^2$ is set entirely by the ones digit of $n$.
That means the ones digits of squares repeat every $10$ terms.
Square the digits $1$ through $10$ and keep each ones digit: $1,4,9,6,5,6,9,4,1,0$.
The next square, $11^2=121$, ends in $1$ again, so this block of ten repeats forever.

$$1^2\to1,\;2^2\to4,\;3^2\to9,\;4^2\to6,\;5^2\to5,\;6^2\to6,\;7^2\to9,\;8^2\to4,\;9^2\to1,\;10^2\to0$$

💡 Only the last digit of a number affects the last digit of its square, so the endings must cycle.

#7 Identify Subproblems 4.NBT.B.4 Step 2

Add one full block

Add the ten ones digits in a single block to see what each complete cycle contributes: $1+4+9+6+5+6+9+4+1+0$.
Pairing $1+9$, $4+6$, $9+1$, $6+4$ gives four tens, plus the leftover $5$, for $45$.
Every full block of ten squares adds $45$ to the running total.

$$1+4+9+6+5+6+9+4+1+0 = 45$$

💡 If a chunk repeats, you only need the sum of one chunk.

#7 Identify Subproblems 5.NBT.B.6 Step 3

Count the whole blocks in 2025 terms

Split the $2025$ terms into groups of $10$.
Dividing, $2025\div10=202$ with a remainder of $5$, so $2025=202\times10+5$.
That is $202$ complete blocks of ten, with $5$ terms left over at the end.

$$2025 = 202\times10 + 5$$

💡 The digits left of the ones place, $202$, count how many complete tens fit inside $2025$.

#7 Identify Subproblems 5.NBT.B.5 Step 4

Total from the full blocks

Each of the $202$ complete blocks contributes $45$, so the full blocks together give $202\times45$.
Compute it as $200\times45+2\times45=9000+90=9090$.

$$202\times45 = 9000+90 = 9090$$

💡 Multiplying the block sum by the number of blocks handles all the repeats at once.

#7 Identify Subproblems 4.NBT.B.4 Step 5

Add the leftover five terms

The $5$ leftover terms correspond to $n=2021,2022,2023,2024,2025$, whose ones digits match the start of the block: $1,4,9,6,5$.
These add to $1+4+9+6+5=25$.
Add this to the block total: $9090+25=9115$.
That is choice (D).
The trap value $9090$ in (C) is exactly what you get if you forget the $5$ leftover terms.

$$9090 + (1+4+9+6+5) = 9090 + 25 = 9115 \;\Rightarrow\; \textbf{(D)}$$

💡 The leftover terms just restart the same block, so their digits are the block's first few.

[1] #5 4.NBT.B.5 The ones digit of a product depends only on the ones digits of the factors, so t

[2] #7 4.NBT.B.4 Add the ten ones digits in a single block to see what each complete cycle contri

[3] #7 5.NBT.B.6 Split the $2025$ terms into groups of $10$. Dividing, $2025\div10=202$ with a re

[4] #7 5.NBT.B.5 Each of the $202$ complete blocks contributes $45$, so the full blocks together

[5] #7 4.NBT.B.4 The $5$ leftover terms correspond to $n=2021,2022,2023,2024,2025$, whose ones di

Review

Reasonableness: A rough estimate confirms the size: the average ones digit is $\frac{45}{10}=4.5$, and $4.5\times2025\approx9112$, right beside $9115$. The total must also end in a $5$: $202\times45$ ends in $0$ and the leftover $25$ ends in $5$, so the sum ends in $5$ — matching (D) $9115$ and ruling out (B), (C), (E). Choice (C) $9090$ is the answer only if you drop the last $5$ terms, and (A) $9025$ comes from miscounting the blocks, so (D) is the consistent result.

Alternative: Instead of splitting off the leftover, note that the digits for $n=1$ to $5$ ($1,4,9,6,5$ summing to $25$) and for $n=6$ to $10$ ($6,9,4,1,0$ summing to $20$) make each half-block. There are $202$ full blocks ($202\times45=9090$) and then one extra first-half ($25$), giving $9090+25=9115$ again.

CCSS standards used (min grade 5)

4.NBT.B.5 Multiply whole numbers using strategies based on place value (Reasoning that the ones digit of $n^2$ depends only on the ones digit of $n$, so the ones digits repeat every ten squares.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Summing one block to $45$, summing the leftover digits to $25$, and combining $9090+25=9115$.)
5.NBT.B.6 Find whole-number quotients and remainders (Dividing $2025$ by $10$ to get $202$ full blocks with a remainder of $5$ terms.)
5.NBT.B.5 Fluently multiply multi-digit whole numbers using the standard algorithm (Computing $202\times45=9090$, the total contributed by all the full blocks.)

⭐ Ones digits of squares repeat every ten numbers, so count the full blocks, multiply by one block's sum, then add the few leftovers.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Find the repeating block of ten

Add one full block

Count the whole blocks in 2025 terms

Total from the full blocks

Add the leftover five terms

Review

CCSS standards used (min grade 5)

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