Problem-Solving StrategyDifficulty 20-90208 tagged problems

Bounding

Bounding is the habit of trapping a quantity between upper and lower estimates until only one answer remains. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$L \le x \le U \text{ with } L, U \text{ forcing a unique integer (or comparison)} \;\Rightarrow\; x \text{ is pinned down.}$$

Trap an unknown quantity between a lower and an upper estimate; squeeze the bounds tight enough that only one value can survive between them.

Why It Works

1
Produce a lower bound $L \le x$ and an upper bound $x \le U$ by estimating or comparing with simpler quantities.
2
Tighten the estimates until the interval $[L, U]$ contains only one candidate of the required type — often exactly one integer.
3
If $x$ must be an integer and $L, U$ are consecutive integers (or $U - L < 1$), the value is forced.
4
A common trick: add a tiny conjugate term whose size you can bound to convert an irrational expression into an integer minus a known small amount.

Worked Examples

Example 1

Find $\left\lfloor (\sqrt{2}+1)^6 \right\rfloor$.

1
Let $a = (\sqrt2+1)^2 = 3 + 2\sqrt2$ and $b = (\sqrt2-1)^2 = 3 - 2\sqrt2$, so $a + b = 6$ and $ab = (3)^2 - (2\sqrt2)^2 = 9 - 8 = 1$.
2
Then $(\sqrt2+1)^6 + (\sqrt2-1)^6 = a^3 + b^3 = (a+b)^3 - 3ab(a+b) = 6^3 - 3\cdot 1 \cdot 6 = 216 - 18 = 198$, an integer.
3
Now bound the conjugate: $0 < \sqrt2 - 1 < 1$, so $0 < (\sqrt2-1)^6 < 1$.
4
Therefore $(\sqrt2+1)^6 = 198 - (\sqrt2-1)^6$ lies strictly between $197$ and $198$, so its floor is $197$.

Answer:

$197$

Contest level

How many digits does $2^{40}$ have? (Use $\log_{10} 2 \approx 0.30103$.)

1
A positive integer $N$ has exactly $\lfloor \log_{10} N \rfloor + 1$ digits, so bound $\log_{10}\!\big(2^{40}\big) = 40\log_{10} 2$.
2
Estimate: $40 \times 0.30103 = 12.0412$, so $12 < \log_{10}\!\big(2^{40}\big) < 13$.
3
Hence $\lfloor \log_{10}\!\big(2^{40}\big) \rfloor = 12$, and the digit count is $12 + 1 = 13$.
4
The bound pins it down: $10^{12} < 2^{40} < 10^{13}$, so $2^{40}$ has $13$ digits (indeed $2^{40} = 1{,}099{,}511{,}627{,}776$).

Answer:

$13$ digits.

Olympiad / Challenge

Prove that for every integer $n \ge 1$, the sum $\displaystyle\sum_{k=1}^{n} \frac{1}{k^2} = 1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2}$ is less than $2$.

1
Bound each term for $k \ge 2$ by a larger, telescoping quantity: $\frac{1}{k^2} < \frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k}$.
2
Sum that bound from $k = 2$ to $n$: $\sum_{k=2}^{n}\Big(\frac{1}{k-1} - \frac{1}{k}\Big) = 1 - \frac{1}{n}$ by telescoping.
3
Add back the $k = 1$ term: $\sum_{k=1}^{n} \frac{1}{k^2} < 1 + \Big(1 - \frac{1}{n}\Big) = 2 - \frac{1}{n}$ for $n \ge 2$ (and the sum equals $1 < 2$ when $n = 1$).
4
Since $2 - \frac{1}{n} < 2$ for every $n$, the partial sums are bounded above by $2$ for all $n$.

Answer:

$\displaystyle\sum_{k=1}^{n} \frac{1}{k^2} < 2 - \frac{1}{n} < 2$ for all $n \ge 2$ (and $< 2$ for $n = 1$ as well).

Going Deeper

Bounding comes in two modes: squeezing to a single integer (floors, digit counts, $U - L < 1$) and bounding to prove an inequality or convergence (telescoping a sum below a finite limit) — both replace an exact unknown with a tight estimate.
It is everywhere on the AIME and olympiads: floor/ceiling problems, 'how many integers satisfy…', estimates of sums and products, and the conjugate trick $(\sqrt2+1)^n + (\sqrt2-1)^n \in \mathbb{Z}$ that converts an irrational power into an integer minus a tiny remainder.
Pitfall: keep track of strict versus non-strict inequalities. To force a unique integer you need the bounds to exclude the endpoints (e.g. $0 < (\sqrt2-1)^6 < 1$ strictly, so the floor is $197$ and not $198$); a sloppy $\le$ at the wrong place lets two values survive.

Spot the Signal

Look for problems where the key step is trapping a quantity between upper and lower estimates until only one answer remains.
You can describe the hard part as trapping a quantity between upper and lower estimates until only one answer remains, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the problem-solving bottleneck that calls for bounding, then rewrite the givens around it.
Name the relation that makes Bounding legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Bounding just because the surface looks familiar; verify the required condition first.
Applying Bounding because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let bounding reveal the strategic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is trapping a quantity between upper and lower estimates until only one answer remains.

Back to techniques

Problem-Solving StrategyDifficulty 20-90208 tagged problems

Bounding

Practice this technique Match my level

The Key Result

$$L \le x \le U \text{ with } L, U \text{ forcing a unique integer (or comparison)} \;\Rightarrow\; x \text{ is pinned down.}$$

Trap an unknown quantity between a lower and an upper estimate; squeeze the bounds tight enough that only one value can survive between them.

Why It Works

1
Produce a lower bound $L \le x$ and an upper bound $x \le U$ by estimating or comparing with simpler quantities.
2
Tighten the estimates until the interval $[L, U]$ contains only one candidate of the required type — often exactly one integer.
3
If $x$ must be an integer and $L, U$ are consecutive integers (or $U - L < 1$), the value is forced.
4
A common trick: add a tiny conjugate term whose size you can bound to convert an irrational expression into an integer minus a known small amount.

Worked Examples

Example 1

Find $\left\lfloor (\sqrt{2}+1)^6 \right\rfloor$.

1
Let $a = (\sqrt2+1)^2 = 3 + 2\sqrt2$ and $b = (\sqrt2-1)^2 = 3 - 2\sqrt2$, so $a + b = 6$ and $ab = (3)^2 - (2\sqrt2)^2 = 9 - 8 = 1$.
2
Then $(\sqrt2+1)^6 + (\sqrt2-1)^6 = a^3 + b^3 = (a+b)^3 - 3ab(a+b) = 6^3 - 3\cdot 1 \cdot 6 = 216 - 18 = 198$, an integer.
3
Now bound the conjugate: $0 < \sqrt2 - 1 < 1$, so $0 < (\sqrt2-1)^6 < 1$.
4
Therefore $(\sqrt2+1)^6 = 198 - (\sqrt2-1)^6$ lies strictly between $197$ and $198$, so its floor is $197$.

Answer:

$197$

Contest level

How many digits does $2^{40}$ have? (Use $\log_{10} 2 \approx 0.30103$.)

1
A positive integer $N$ has exactly $\lfloor \log_{10} N \rfloor + 1$ digits, so bound $\log_{10}\!\big(2^{40}\big) = 40\log_{10} 2$.
2
Estimate: $40 \times 0.30103 = 12.0412$, so $12 < \log_{10}\!\big(2^{40}\big) < 13$.
3
Hence $\lfloor \log_{10}\!\big(2^{40}\big) \rfloor = 12$, and the digit count is $12 + 1 = 13$.
4
The bound pins it down: $10^{12} < 2^{40} < 10^{13}$, so $2^{40}$ has $13$ digits (indeed $2^{40} = 1{,}099{,}511{,}627{,}776$).

Answer:

$13$ digits.

Olympiad / Challenge

Prove that for every integer $n \ge 1$, the sum $\displaystyle\sum_{k=1}^{n} \frac{1}{k^2} = 1 + \frac{1}{4} + \frac{1}{9} + \cdots + \frac{1}{n^2}$ is less than $2$.

1
Bound each term for $k \ge 2$ by a larger, telescoping quantity: $\frac{1}{k^2} < \frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k}$.
2
Sum that bound from $k = 2$ to $n$: $\sum_{k=2}^{n}\Big(\frac{1}{k-1} - \frac{1}{k}\Big) = 1 - \frac{1}{n}$ by telescoping.
3
Add back the $k = 1$ term: $\sum_{k=1}^{n} \frac{1}{k^2} < 1 + \Big(1 - \frac{1}{n}\Big) = 2 - \frac{1}{n}$ for $n \ge 2$ (and the sum equals $1 < 2$ when $n = 1$).
4
Since $2 - \frac{1}{n} < 2$ for every $n$, the partial sums are bounded above by $2$ for all $n$.

Answer:

$\displaystyle\sum_{k=1}^{n} \frac{1}{k^2} < 2 - \frac{1}{n} < 2$ for all $n \ge 2$ (and $< 2$ for $n = 1$ as well).

Going Deeper

Bounding comes in two modes: squeezing to a single integer (floors, digit counts, $U - L < 1$) and bounding to prove an inequality or convergence (telescoping a sum below a finite limit) — both replace an exact unknown with a tight estimate.
It is everywhere on the AIME and olympiads: floor/ceiling problems, 'how many integers satisfy…', estimates of sums and products, and the conjugate trick $(\sqrt2+1)^n + (\sqrt2-1)^n \in \mathbb{Z}$ that converts an irrational power into an integer minus a tiny remainder.
Pitfall: keep track of strict versus non-strict inequalities. To force a unique integer you need the bounds to exclude the endpoints (e.g. $0 < (\sqrt2-1)^6 < 1$ strictly, so the floor is $197$ and not $198$); a sloppy $\le$ at the wrong place lets two values survive.

Spot the Signal

Look for problems where the key step is trapping a quantity between upper and lower estimates until only one answer remains.
You can describe the hard part as trapping a quantity between upper and lower estimates until only one answer remains, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the problem-solving bottleneck that calls for bounding, then rewrite the givens around it.
Name the relation that makes Bounding legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Bounding just because the surface looks familiar; verify the required condition first.
Applying Bounding because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let bounding reveal the strategic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is trapping a quantity between upper and lower estimates until only one answer remains.