AMC 10 · 2009 · #5
Grade 5 number-theoryWhat is the sum of the digits of the square of 111111111?
Pick an answer.
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.
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.
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) $81$
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
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.
💡 Shrinking the number keeps the same structure while making the multiplication small enough to do by hand.
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.
💡 Each extra $1$ just adds one more rung to a staircase that goes up to the peak and mirrors back down.
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 $1$.
💡 Nine $1$s means the middle digit is $9$, so the staircase reaches its tallest possible single-digit peak.
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).
💡 The digits mirror around the peak, so counting up to $9$ and back down folds into the neat square $9^2$.
5.NBT.B.5 Start with easier numbers that look the same but are shorter. Square $11$, then 4.OA.C.5 Line the results up: $1^2=1$, $11^2=121$, $111^2=12321$, $1111^2=1234321$. A rep 4.OA.C.5 Here $n=9$, exactly the last case before carrying would begin. So the peak digit 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.5Fluently multiply multi-digit whole numbers (Squaring the small repunits $11$, $111$, $1111$ to generate the first data points.)4.OA.C.5Generate 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.2Fluently 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 $9^2=81$.
⭐ 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 $9^2=81$.
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