Invariants is the habit of finding a quantity that stays unchanged through moves, operations, or transformations. 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.
$$\text{A quantity } I \text{ is invariant if every allowed move leaves } I \text{ unchanged.}$$
Find a quantity that never changes under the allowed moves; states with different values of it can never be reached from one another.
Why It Works
1
Identify the moves (operations) the problem permits.
2
Search for a quantity — often a parity, a sum modulo $n$, or a coloring count — that is preserved by every single move.
3
If such an invariant $I$ exists, then $I$ has the same value at the start and at any reachable state.
4
Therefore any target state with a different value of $I$ is unreachable — an impossibility proof with no case-bashing.
Worked Examples
Example 1
Numbers $1, 2, \ldots, 10$ are on a board. A move erases two numbers $a, b$ and writes $a + b$. Can repeated moves end with the single number $54$?
1
Each move replaces $a, b$ by $a + b$, so the sum of all numbers on the board never changes — the total sum is invariant.
2
The initial sum is $1 + 2 + \cdots + 10 = \tfrac{10 \cdot 11}{2} = 55$.
3
Any final single number must equal that invariant sum, $55$.
4
Since $54 \ne 55$, ending at $54$ is impossible.
Answer:
No — the sum is invariant at $55$, so only $55$ can remain.
Warm-up
A jar holds $7$ black and $8$ white beads. Repeatedly remove two beads: if they are the same color, add back one black bead; if different colors, add back one white bead. After many moves one bead remains. What color is it?
1
Track the parity of the number of WHITE beads. Two-black removal: blacks $-2$, whites $0$ — white count unchanged. Two-white removal: whites $-2$, then $+1$ black, so whites drop by $2$. Mixed removal: one white removed, then a white added back, so whites unchanged.
2
In every move the number of white beads changes by $0$ or $-2$, so its PARITY never changes — an invariant.
3
Initially there are $8$ white beads, which is even.
4
The final single bead leaves either $0$ whites (even) or $1$ white (odd). To preserve even parity it must be $0$ whites, so the last bead is black.
Answer:
Black — the parity of the white-bead count is invariant (even), forcing $0$ whites at the end.
Contest level
The numbers $+1$ and $-1$ are written in the cells of a $4 \times 4$ grid (sixteen entries). A move flips the sign of every entry in a chosen row or a chosen column. Can a grid of all $+1$ be turned into a grid with exactly one $-1$ and fifteen $+1$?
1
Consider the PRODUCT of all sixteen entries. Flipping a row negates $4$ entries, multiplying the product by $(-1)^4 = +1$; flipping a column likewise multiplies by $(-1)^4 = +1$.
2
So every move leaves the product of all entries unchanged — the product is an invariant.
3
The all-$+1$ grid has product $+1$.
4
A grid with exactly one $-1$ and fifteen $+1$ has product $(-1) \cdot 1^{15} = -1 \ne +1$, so it can never be reached.
Answer:
No — the product of all entries is invariant at $+1$, but a single $-1$ would make the product $-1$.
Olympiad / Challenge
Three piles hold $11$, $12$, and $13$ chips. A move takes one chip from each of two piles and adds two chips to the third. Can repeated moves make all three piles equal?
1
Track the pile sizes modulo $3$. A move sends $(a, b, c) \to (a - 1, b - 1, c + 2)$ (or a permutation). Each entry changes by $-1$ or $+2$, and $-1 \equiv +2 \pmod 3$, so every residue increases by $2 \pmod 3$.
2
Because all three residues shift by the SAME amount each move, the pairwise DIFFERENCES of residues are invariant mod $3$.
3
Initial residues: $11 \equiv 2$, $12 \equiv 0$, $13 \equiv 1 \pmod 3$ — all three are distinct, so the pairwise differences are nonzero mod $3$.
4
Equal piles would require all three residues equal (pairwise differences $\equiv 0$), but those differences are invariant and start nonzero, so the piles can never coincide. (The total $11+12+13 = 36$ is divisible by $3$, so the sum alone does not forbid equality — the mod-$3$ difference invariant is what kills it.)
Answer:
No — the pairwise differences of pile sizes mod $3$ are invariant and start all distinct, so the piles can never be made equal.
Going Deeper
The general recipe: list the moves, then hunt for a quantity preserved (or constrained to a small orbit) by every move — common choices are a sum or product, a parity, a value mod $m$, a coloring count (e.g. on a checkerboard), or a more exotic 'monovariant' that only increases. If the start and target disagree on the invariant, the target is unreachable.
Invariants dominate olympiad 'is it possible to reach...' and 'prove you cannot...' problems: chip-firing, coin/token games, tilings (a defective chessboard cannot be domino-tiled because dominoes preserve the black-minus-white count), and sign-flipping grids. A monovariant (a quantity that strictly decreases) instead proves a process must terminate.
Pitfall: an invariant proves IMPOSSIBILITY, never possibility — matching invariants is necessary but not sufficient, so you still must exhibit an explicit sequence of moves to prove a target IS reachable. The other trap is claiming a quantity is invariant without checking EVERY move type; one overlooked move that changes it collapses the whole argument.
Spot the Signal
Look for problems where the key step is finding a quantity that stays unchanged through moves, operations, or transformations.
You can describe the hard part as finding a quantity that stays unchanged through moves, operations, or transformations, 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 counting object that calls for invariants, then rewrite the givens around it.
Name the relation that makes Invariants 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 Invariants just because the surface looks familiar; verify the required condition first.
Applying Invariants 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 invariants reveal the counting structure; then compute only what remains.
Try it on:
Practice a contest problem where the key step is finding a quantity that stays unchanged through moves, operations, or transformations.