AMC 10 · 2003 · #7
Grade 8 arithmeticThe symbolism ⌊x⌋ denotes the largest integer not exceeding x. For example, ⌊3⌋=3, and ⌊9/2⌋=4. Compute
⌊1⌋+⌊2⌋+⌊3⌋+⋯+⌊16⌋.
Pick an answer.
AMC 10 2003 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: The symbol $\lfloor x \rfloor$ means the largest integer that is not bigger than $x$ (round down). Add up $\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \cdots + \lfloor \sqrt{16} \rfloor$, one term for each whole number from $1$ to $16$, and choose which value the total equals.
Givens: $\lfloor x \rfloor$ is $x$ rounded down to a whole number, e.g. $\lfloor 3 \rfloor = 3$ and $\lfloor 9/2 \rfloor = 4$; The sum runs over every integer $n$ from $1$ to $16$; Each term is $\lfloor \sqrt{n} \rfloor$; Answer choices: (A) $35$, (B) $38$, (C) $40$, (D) $42$, (E) $136$
Unknowns: The value of the whole sum $\lfloor \sqrt{1} \rfloor + \cdots + \lfloor \sqrt{16} \rfloor$
Understand
Restated: The symbol $\lfloor x \rfloor$ means the largest integer that is not bigger than $x$ (round down). Add up $\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \cdots + \lfloor \sqrt{16} \rfloor$, one term for each whole number from $1$ to $16$, and choose which value the total equals.
Givens: $\lfloor x \rfloor$ is $x$ rounded down to a whole number, e.g. $\lfloor 3 \rfloor = 3$ and $\lfloor 9/2 \rfloor = 4$; The sum runs over every integer $n$ from $1$ to $16$; Each term is $\lfloor \sqrt{n} \rfloor$; Answer choices: (A) $35$, (B) $38$, (C) $40$, (D) $42$, (E) $136$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
Writing out all sixteen square roots and rounding each one is slow and error-prone, so Tool #5 (Look for a Pattern) is the spine: $\lfloor \sqrt{n} \rfloor$ does not change until $n$ reaches the next perfect square, so the terms come in flat blocks of equal value. Tool #7 (Identify Subproblems) then splits the sum into those blocks — a block of $1$s, a block of $2$s, a block of $3$s, and the single $4$ — each an easy 'value times how many' piece. Tool #3 (Eliminate Possibilities) guards the finish: the trap choice (E) $136$ is what you get by adding $1+2+\cdots+16$ (forgetting the square root entirely), and other near-misses come from miscounting a block, so grouping keeps the count honest.
Execute — Answer: B
8.NS.A.2 Step 1 See where the value stays flat
- $\lfloor \sqrt{n} \rfloor$ asks: what is the biggest whole number whose square is not larger than $n$?
- That answer stays the same as $n$ climbs, and only steps up by one when $n$ reaches the next perfect square.
- So $\lfloor \sqrt{n} \rfloor = k$ for every $n$ from $k^2$ up to just below $(k+1)^2$.
💡 The rounded-down square root can only change at a perfect square, so between squares it holds steady.
8.EE.A.2 Step 2 Mark the block boundaries with perfect squares
- The perfect squares up to $16$ are $1=1^2$, $4=2^2$, $9=3^2$, and $16=4^2$.
- So the value is $1$ for $n=1,2,3$ (up to just before $4$), then $2$ for $n=4,5,6,7,8$ (up to just before $9$), then $3$ for $n=9,\ldots,15$ (up to just before $16$), and finally $4$ at $n=16$ itself.
💡 The perfect squares are the exact spots where the rounded square root jumps up.
8.NS.A.2 Step 3 Count the terms in each block
- Each block runs from one perfect square up to just below the next, so its length is the gap between consecutive squares: $4-1=3$ terms equal to $1$, $9-4=5$ terms equal to $2$, $16-9=7$ terms equal to $3$, and the single term $n=16$ equal to $4$.
- The block sizes $3, 5, 7$ are the odd numbers, which is the pattern $(k+1)^2 - k^2 = 2k+1$.
💡 How many numbers sit between two squares is just the distance between those squares.
3.OA.A.1 Step 4 Multiply each block and add
- Each block is a stack of equal terms, so its total is the value times how many there are: $1\times3=3$, $2\times5=10$, $3\times7=21$, and $4\times1=4$.
- Adding the four block totals gives $3+10+21+4=38$, which matches choice (B).
💡 A block of identical numbers adds up to one number times how many copies there are.
8.NS.A.2 $\lfloor \sqrt{n} \rfloor$ asks: what is the biggest whole number whose square i 8.EE.A.2 The perfect squares up to $16$ are $1=1^2$, $4=2^2$, $9=3^2$, and $16=4^2$. So t 8.NS.A.2 Each block runs from one perfect square up to just below the next, so its length 3.OA.A.1 Each block is a stack of equal terms, so its total is the value times how many t Review
Reasonableness: The sixteen terms are all between $1$ and $4$, so the total must sit between $16\times1=16$ and $16\times4=64$; the answer $38$ lands squarely in that range, while (E) $136$ is far too big and is exactly $1+2+\cdots+16$ — the trap you fall into by forgetting the square root. A quick recount of the blocks $3+5+7+1=16$ terms confirms every number from $1$ to $16$ is used exactly once, so no term was dropped or double-counted, and $38$ is choice (B).
Alternative: Tool #2 (Make a Systematic List): just write the sixteen rounded roots in order — $1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,4$ — and add straight across. It is more writing, but it lands on the same $38$ and doubles as a check that the blocks were counted right.
CCSS standards used (min grade 8)
8.NS.A.2Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (Seeing that $\lfloor \sqrt{n} \rfloor$ holds the value $k$ for every $n$ from $k^2$ up to just below $(k+1)^2$, and counting how many such $n$ there are.)8.EE.A.2Use square root and cube root symbols to represent solutions to equations, and evaluate square roots of small perfect squares (Pinning the block boundaries at the perfect squares $1, 4, 9, 16 = 1^2, 2^2, 3^2, 4^2$.)3.OA.A.1Interpret products of whole numbers, e.g., interpret 5 x 7 as the total number of objects in 5 groups of 7 objects each (Turning each flat block into value times count ($1\cdot3, 2\cdot5, 3\cdot7, 4\cdot1$) and adding the results.)
⭐ A rounded-down square root only jumps at perfect squares, so group the sum into flat blocks between the squares and add value times count.
⭐ A rounded-down square root only jumps at perfect squares, so group the sum into flat blocks between the squares and add value times count.
More like this
Same archetype — closest grade level first.