AMC 10 · 2009 · #5

Easy mode Grade 5

Problem

Take the number 111,111,111 — that is nine 1s in a row. Square it, which means multiply it by itself. Then add up all the digits of the answer. What is that sum?

Pick an answer.

(A)

$\ 18$

(B)

$\ 27$

(C)

$\ 45$

(D)

$\ 63$

(E)

$\ 81$

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

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Toolkit + CCSS Solution

Understand

Restated: Take the number made of nine $1$s, $111111111$, square it, and add up all the digits of the result. Report that digit sum.

Givens: The base number is $111111111$, which is $1$ repeated nine times; We must square it: $111111111^2$; We then add every digit of that square together; Answer choices: (A) $18$, (B) $27$, (C) $45$, (D) $63$, (E) $81$

Unknowns: The sum of the digits of $111111111^2$

Understand

Restated: Take the number made of nine $1$s, $111111111$, square it, and add up all the digits of the result. Report that digit sum.

Plan

Primary tool: #5 Look for a Pattern

Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems

Squaring $111111111$ directly is heavy, so first shrink the problem (Tool #9): square the small repunits $11$, $111$, $1111$ where the arithmetic is easy. Those small squares reveal a clean shape — a number that counts up then back down — which is exactly what Tool #5 (Look for a Pattern) is for. Once the pattern gives the full square, adding its digits splits into a tidy subproblem (Tool #7): a run of $1$ up to $9$ and back down, which sums to a perfect square.

Execute — Answer: E

#9 Solve an Easier Related Problem 5.NBT.B.5 Step 1

Square the small repunits

Start with easier numbers that look the same but are shorter.
Square $11$, then $111$, then $1111$ using ordinary multiplication.
Each answer is a palindrome — it reads the same forwards and backwards.

$$11^2=121,\quad 111^2=12321,\quad 1111^2=1234321$$

💡 Shrinking the number keeps the same structure while making the multiplication small enough to do by hand.

#5 Look for a Pattern 4.OA.C.5 Step 2

Spot the count-up-count-down pattern

Line the results up: $1^2=1$, $11^2=121$, $111^2=12321$, $1111^2=1234321$.
A repunit with $n$ ones, when squared, gives the digits climbing $1,2,3,\dots$ up to $n$ and then stepping back down $\dots,3,2,1$.
The count of $1$s becomes the peak digit in the middle.
This works cleanly as long as $n\le 9$, because no digit reaches $10$ to force a carry.

$$1,\;121,\;12321,\;1234321,\;\dots \;\Rightarrow\; \text{climb } 1\!\to\! n \text{ then back to } 1$$

💡 Each extra $1$ just adds one more rung to a staircase that goes up to the peak and mirrors back down.

#5 Look for a Pattern 4.OA.C.5 Step 3

Apply the pattern to nine 1s

Here $n=9$, exactly the last case before carrying would begin.
So the peak digit is $9$, and the square counts up $1$ through $9$ and back down to $1$.

$$111111111^2 = 12345678987654321$$

💡 Nine $1$s means the middle digit is $9$, so the staircase reaches its tallest possible single-digit peak.

#7 Identify Subproblems 3.NBT.A.2 Step 4

Add the digits

Now sum the digits of $12345678987654321$.
Group them as the climb up to $9$ and the fall back down: $(1+2+\cdots+8) + 9 + (8+\cdots+2+1)$.
That is $36 + 9 + 36 = 81$.
A quicker view: the whole run $1+2+\cdots+9+\cdots+2+1$ equals $9^2=81$.
So the digit sum is $81$, which is choice (E).

$$1+2+\cdots+9+\cdots+2+1 = 9^2 = 81 \;\textbf{(E)}$$

💡 The digits mirror around the peak, so counting up to $9$ and back down folds into the neat square $9^2$.

[1] #9 5.NBT.B.5 Start with easier numbers that look the same but are shorter. Square $11$, then

[2] #5 4.OA.C.5 Line the results up: $1^2=1$, $11^2=121$, $111^2=12321$, $1111^2=1234321$. A rep

[3] #5 4.OA.C.5 Here $n=9$, exactly the last case before carrying would begin. So the peak digit

[4] #7 3.NBT.A.2 Now sum the digits of $12345678987654321$. Group them as the climb up to $9$ and

Review

Reasonableness: The answer $81$ is one of the choices, and it equals $9^2$ — the square of the number of $1$s — which is a memorable, clean result. A sanity check on scale: the square $12345678987654321$ has $17$ digits, and an average digit near $5$ would predict a sum around $85$, right where $81$ sits. Smaller cases confirm the rule: $111^2=12321$ has digit sum $9=3^2$, and $1111^2=1234321$ has digit sum $16=4^2$, so nine $1$s giving $9^2=81$ fits the family exactly.

Alternative: Instead of the pattern, note that the digit sum of $n^2$ equals $n^2 \bmod 9$ shifted by $9$s (casting out nines). Since $111111111$ is nine $1$s, its digit sum is $9$, so it is divisible by $9$; its square is divisible by $81$ and has digit sum a multiple of $9$. Among the choices only $18,27,45,63,81$ are multiples of $9$ — all of them — so this narrows nothing by itself, but combined with the size estimate ($17$ digits, sum near $80$) it points to $81$. The pattern method is cleaner and exact.

CCSS standards used (min grade 5)

5.NBT.B.5 Fluently multiply multi-digit whole numbers (Squaring the small repunits $11$, $111$, $1111$ to generate the first data points.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Recognizing the count-up-count-down palindrome rule and extending it to nine $1$s: $12345678987654321$.)
3.NBT.A.2 Fluently add and subtract within 1000 (Adding the digits $1+2+\cdots+9+\cdots+2+1$ to get $81$.)

⭐ Squaring a string of $1$s makes the digits count up then back down, so nine $1$s square to $12345678987654321$ whose digits add to $9^2=81$.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Square the small repunits

Spot the count-up-count-down pattern

Apply the pattern to nine 1s

Add the digits

Review

CCSS standards used (min grade 5)

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