p-adic Valuations is the habit of counting the exponent of a prime inside an integer to sharpen divisibility arguments. 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.
So $v_5(100!) = 24$. We have $v_2(100!) > 24$ for free: term by term $\lfloor 100/2^i \rfloor \ge \lfloor 100/5^i \rfloor$ (multiples of $2^i$ are at least as common as multiples of $5^i$), so $v_2(n!) \ge v_5(n!)$ always. Hence the number of $10$s is $\min(v_2, v_5) = v_5(100!) = 24$.
4
Each factor of $10$ gives one trailing zero.
Answer:
$24$
Warm-up
Find $v_2(50!)$, the exponent of $2$ in $50!$.
1
Apply Legendre's formula with $p = 2$, $n = 50$: $v_2(50!) = \sum_{i \ge 1} \lfloor 50/2^i \rfloor$.
There is a slick check via the digit-sum form $v_p(n!) = \frac{n - s_p(n)}{p-1}$: in base $3$, $100 = 10201_3$ with digit sum $s_3(100) = 1+0+2+0+1 = 4$, so $v_3(100!) = \frac{100 - 4}{3 - 1} = \frac{96}{2} = 48$. ✓
Answer:
$48$
Olympiad / Challenge
Prove that the central binomial coefficient $\binom{2n}{n}$ is even for every positive integer $n$.
1
We compute $v_2\!\left(\binom{2n}{n}\right)$ and show it is at least $1$. Use Legendre on $\binom{2n}{n} = \frac{(2n)!}{n!\,n!}$: $v_2 = \sum_{i \ge 1}\left(\lfloor 2n/2^i \rfloor - 2\lfloor n/2^i \rfloor\right)$.
2
By Kummer's theorem this sum equals the number of carries when adding $n + n$ in base $2$. Equivalently, in terms of binary digit sums, $v_2\!\left(\binom{2n}{n}\right) = 2 s_2(n) - s_2(2n)$.
3
Doubling shifts the binary representation one place left and appends a $0$, so $s_2(2n) = s_2(n)$. Hence $v_2\!\left(\binom{2n}{n}\right) = 2 s_2(n) - s_2(n) = s_2(n)$.
4
For $n \ge 1$ the digit sum $s_2(n) \ge 1$, so $v_2\!\left(\binom{2n}{n}\right) = s_2(n) \ge 1$, meaning $2 \mid \binom{2n}{n}$. (It also shows $\binom{2n}{n}$ is a power of $2$ times an odd number with exactly $s_2(n)$ factors of $2$.)
Answer:
$v_2\!\left(\binom{2n}{n}\right) = s_2(n) \ge 1$, so $\binom{2n}{n}$ is always even.
Going Deeper
Generalization: Legendre's formula has the compact digit-sum form $v_p(n!) = \frac{n - s_p(n)}{p - 1}$, where $s_p(n)$ is the sum of the base-$p$ digits of $n$. For binomial coefficients, Kummer's theorem upgrades this: $v_p\!\left(\binom{m+n}{m}\right)$ equals the number of carries when adding $m + n$ in base $p$.
Where it appears: counting trailing zeros of $n!$ (count $v_5$), proving binomial coefficients are integers, locating the exact power of a prime in factorials/binomials/multinomials, and as the engine inside Lifting the Exponent and many divisibility proofs.
Pitfall: the sum is finite — stop once $p^i > n$ (all later terms are $0$); a common slip is to keep adding or to miscount $\lfloor n/p^i \rfloor$. Also note trailing zeros of $n!$ are governed by $\min(v_2, v_5) = v_5$ because fives are always scarcer than twos — do NOT add $v_2$ and $v_5$.
Spot the Signal
Look for problems where the key step is counting the exponent of a prime inside an integer to sharpen divisibility arguments.
You can describe the hard part as counting the exponent of a prime inside an integer to sharpen divisibility arguments, 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 integer constraint that calls for p-adic valuations, then rewrite the givens around it.
Name the relation that makes p-adic Valuations 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 p-adic Valuations just because the surface looks familiar; verify the required condition first.
Applying p-adic Valuations 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 p-adic valuations reveal the integer structure; then compute only what remains.
Try it on:
Practice a contest problem where the key step is counting the exponent of a prime inside an integer to sharpen divisibility arguments.